Bernoulli number · Lablypedia

The values of the first 20 Bernoulli numbers are given in the adjacent table. Two conventions are used in the literature, denoted here by $B_{n}^{-}$ and $B_{n}^{+}$ ; they differ only for n = 1, where $B_{1}^{-} = - 1 / 2$ and $B_{1}^{+} = + 1 / 2$ . For every odd n > 1, B_n = 0. For every even n > 0, B_n is negative if n is divisible by 4 and positive otherwise. The Bernoulli numbers are special values of the Bernoulli polynomials $B_{n} (x)$ , with $B_{n}^{-} = B_{n} (0)$ and $B_{n}^{+} = B_{n} (1)$ .

The Bernoulli numbers were discovered around the same time by the Swiss mathematician Jacob Bernoulli, after whom they are named, and independently by Japanese mathematician Seki Takakazu. Seki's discovery was posthumously published in 1712 in his work Katsuyō Sanpō; Bernoulli's, also posthumously, in his Ars Conjectandi of 1713. Ada Lovelace's note G on the Analytical Engine from 1842 describes an algorithm for generating Bernoulli numbers with Babbage's machine; it is disputed whether Lovelace or Babbage developed the algorithm. As a result, the Bernoulli numbers have the distinction of being the subject of the first published complex computer program.

Notation

The superscript ± used in this article distinguishes the two sign conventions for Bernoulli numbers. Only the n = 1 term is affected:

B⁻
_n with B⁻
₁ = −⁠1/2⁠ (OEIS: A027641 / OEIS: A027642) is the sign convention prescribed by NIST and many modern textbooks.
B⁺
_n with B⁺
₁ = +⁠1/2⁠ (OEIS: A164555 / OEIS: A027642) was used in the older literature, and (since 2022) by Donald Knuth following Peter Luschny's "Bernoulli Manifesto".

In the formulas below, one can switch from one sign convention to the other with the relation $B_{n}^{+} = (- 1)^{n} B_{n}^{-}$ , or for integer n = 2 or greater, simply ignore it.

Since B_n = 0 for all odd n > 1, and many formulas only involve even-index Bernoulli numbers, a few authors write "B_n" instead of B_2n . This article does not follow that notation.

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History

Early history

The Bernoulli numbers are rooted in the early history of the computation of sums of integer powers, which have been of interest to mathematicians since antiquity.

Methods to calculate the sum of the first n positive integers, the sum of the squares and of the cubes of the first n positive integers were known, but there were no real 'formulas', only descriptions given entirely in words. Among the great mathematicians of antiquity to consider this problem were Pythagoras (c. 572–497 BCE, Greece), Archimedes (287–212 BCE, Italy), Aryabhata (b. 476, India), Al-Karaji (d. 1019, Persia) and Ibn al-Haytham (965–1039, Iraq).

During the late sixteenth and early seventeenth centuries mathematicians made significant progress. In the West Thomas Harriot (1560–1621) of England, Johann Faulhaber (1580–1635) of Germany, Pierre de Fermat (1601–1665) and fellow French mathematician Blaise Pascal (1623–1662) all played important roles.

Thomas Harriot seems to have been the first to derive and write formulas for sums of powers using symbolic notation, but even he calculated only up to the sum of the fourth powers. Johann Faulhaber gave formulas for sums of powers up to the 17th power in his 1631 Academia Algebrae, far higher than anyone before him, but he did not give a general formula.

Blaise Pascal in 1654 proved Pascal's identity relating (n+1)^k+1 to the sums of the pth powers of the first n positive integers for p = 0, 1, 2, ..., k.

The Swiss mathematician Jacob Bernoulli (1654–1705) was the first to realize the existence of a single sequence of constants B₀, B₁, B₂,... which provide a uniform formula for all sums of powers.

The joy Bernoulli experienced when he hit upon the pattern needed to compute quickly and easily the coefficients of his formula for the sum of the cth powers for any positive integer c can be seen from his comment. He wrote:

"With the help of this table, it took me less than half of a quarter of an hour to find that the tenth powers of the first 1000 numbers being added together will yield the sum 91,409,924,241,424,243,424,241,924,242,500."

Bernoulli's result was published posthumously in Ars Conjectandi in 1713. Seki Takakazu independently discovered the Bernoulli numbers and his result was published a year earlier, also posthumously, in 1712. However, Seki did not present his method as a formula based on a sequence of constants.

Bernoulli's formula for sums of powers is the most useful and generalizable formulation to date. The coefficients in Bernoulli's formula are now called Bernoulli numbers, following a suggestion of Abraham de Moivre.

Bernoulli's formula is sometimes called Faulhaber's formula after Johann Faulhaber who found remarkable ways to calculate sum of powers but never stated Bernoulli's formula. According to Knuth a rigorous proof of Faulhaber's formula was first published by Carl Jacobi in 1834. Knuth's in-depth study of Faulhaber's formula concludes:

"Faulhaber never discovered the Bernoulli numbers; i.e., he never realized that a single sequence of constants B₀, B₁, B₂, ... would provide a uniform

\sum n^{m} = \frac{1}{m + 1} (B_{0} n^{m + 1} - (\binom{m + 1}{1}) B_{1} n^{m} + (\binom{m + 1}{2}) B_{2} n^{m - 1} - \dots + (- 1)^{m} (\binom{m + 1}{m}) B_{m} n)

for all sums of powers. He never mentioned, for example, the fact that almost half of the coefficients turned out to be zero after he had converted his formulas for Σ n^m from polynomials in N to polynomials in n."

In the above Knuth meant $B_{1}^{-}$ ; instead using $B_{1}^{+}$ the formula avoids subtraction:

\sum n^{m} = \frac{1}{m + 1} (B_{0} n^{m + 1} + (\binom{m + 1}{1}) B_{1}^{+} n^{m} + (\binom{m + 1}{2}) B_{2} n^{m - 1} + \dots + (\binom{m + 1}{m}) B_{m} n) .

Reconstruction of "Summæ Potestatum"

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width:16.639ex; height:3.509ex;" alt="{\displaystyle \textstyle \int n={\frac {1}{2}}nn+{\frac {1}{2}}n}"/></span><br/><span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt118" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"\\textstyle \\int nn=\\frac{1}{3}n^{3}+\\frac{1}{2}nn+\\frac{1}{6}n"}}'><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle \textstyle \int nn={\frac {1}{3}}n^{3}+{\frac {1}{2}}nn+{\frac {1}{6}}n}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <mstyle displaystyle="false" scriptlevel="0"> <mo>∫<!-- ∫ --></mo> <mi>n</mi> <mi>n</mi> <mo>=</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>3</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>3</mn> </mrow> </msup> <mo>+</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> </mrow> <mi>n</mi> <mi>n</mi> <mo>+</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>6</mn> </mfrac> </mrow> <mi>n</mi> </mstyle> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle \textstyle \int nn={\frac {1}{3}}n^{3}+{\frac {1}{2}}nn+{\frac {1}{6}}n}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/5f90e482de4f088204b165679233069433596398" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.338ex; width:24.981ex; height:3.676ex;" alt="{\displaystyle \textstyle \int nn={\frac {1}{3}}n^{3}+{\frac {1}{2}}nn+{\frac {1}{6}}n}"/></span><br/><span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt119" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"\\textstyle \\int n^{3}=\\frac{1}{4}n^{4}+\\frac{1}{2}n^{3}+\\frac{1}{4}nn"}}'><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle \textstyle \int n^{3}={\frac {1}{4}}n^{4}+{\frac {1}{2}}n^{3}+{\frac {1}{4}}nn}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <mstyle displaystyle="false" scriptlevel="0"> <mo>∫<!-- ∫ --></mo> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>3</mn> </mrow> </msup> <mo>=</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>4</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>4</mn> </mrow> </msup> <mo>+</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>3</mn> </mrow> </msup> <mo>+</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>4</mn> </mfrac> </mrow> <mi>n</mi> <mi>n</mi> </mstyle> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle \textstyle \int n^{3}={\frac {1}{4}}n^{4}+{\frac {1}{2}}n^{3}+{\frac {1}{4}}nn}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/8e7d24e8a34ea317628f094e4205e244af2d8dd7" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.171ex; width:25.695ex; height:3.509ex;" alt="{\displaystyle \textstyle \int n^{3}={\frac {1}{4}}n^{4}+{\frac {1}{2}}n^{3}+{\frac {1}{4}}nn}"/></span><br/><span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt120" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"\\textstyle \\int n^{4}=\\frac{1}{5}n^{5}+\\frac{1}{2}n^{4}+\\frac{1}{3}n^{3}-\\frac{1}{30}n"}}'><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle \textstyle \int n^{4}={\frac {1}{5}}n^{5}+{\frac {1}{2}}n^{4}+{\frac {1}{3}}n^{3}-{\frac {1}{30}}n}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <mstyle displaystyle="false" scriptlevel="0"> <mo>∫<!-- ∫ --></mo> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>4</mn> </mrow> </msup> <mo>=</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>5</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>5</mn> </mrow> </msup> <mo>+</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>4</mn> </mrow> </msup> <mo>+</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>3</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>3</mn> </mrow> </msup> <mo>−<!-- − --></mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>30</mn> </mfrac> </mrow> <mi>n</mi> </mstyle> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle \textstyle \int n^{4}={\frac {1}{5}}n^{5}+{\frac {1}{2}}n^{4}+{\frac {1}{3}}n^{3}-{\frac {1}{30}}n}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/56899d6a5e6c53a8703bfab1ae30abe28c2352ad" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.338ex; 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width:34.286ex; height:3.676ex;" alt="{\displaystyle \textstyle \int n^{5}={\frac {1}{6}}n^{6}+{\frac {1}{2}}n^{5}+{\frac {5}{12}}n^{4}-{\frac {1}{12}}nn}"/></span><br/><span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt122" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"\\textstyle \\int n^{6}=\\frac{1}{7}n^{7}+\\frac{1}{2}n^{6}+\\frac{1}{2}n^{5}-\\frac{1}{6}n^{3}+\\frac{1}{42}n"}}'><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle \textstyle \int n^{6}={\frac {1}{7}}n^{7}+{\frac {1}{2}}n^{6}+{\frac {1}{2}}n^{5}-{\frac {1}{6}}n^{3}+{\frac {1}{42}}n}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <mstyle displaystyle="false" scriptlevel="0"> <mo>∫<!-- ∫ --></mo> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>6</mn> </mrow> </msup> <mo>=</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>7</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>7</mn> </mrow> </msup> <mo>+</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>6</mn> </mrow> </msup> <mo>+</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>5</mn> </mrow> </msup> <mo>−<!-- − --></mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>6</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>3</mn> </mrow> </msup> <mo>+</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>42</mn> </mfrac> </mrow> <mi>n</mi> </mstyle> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle \textstyle \int n^{6}={\frac {1}{7}}n^{7}+{\frac {1}{2}}n^{6}+{\frac {1}{2}}n^{5}-{\frac {1}{6}}n^{3}+{\frac {1}{42}}n}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/3ab32c035830855a9523ec218c1400fe43e20698" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.338ex; width:39.017ex; height:3.676ex;" alt="{\displaystyle \textstyle \int n^{6}={\frac {1}{7}}n^{7}+{\frac {1}{2}}n^{6}+{\frac {1}{2}}n^{5}-{\frac {1}{6}}n^{3}+{\frac {1}{42}}n}"/></span><br/><span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt123" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"\\textstyle \\int n^{7}=\\frac{1}{8}n^{8}+\\frac{1}{2}n^{7}+\\frac{7}{12}n^{6}-\\frac{7}{24}n^{4}+\\frac{1}{12}nn"}}'><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle \textstyle \int n^{7}={\frac {1}{8}}n^{8}+{\frac {1}{2}}n^{7}+{\frac {7}{12}}n^{6}-{\frac {7}{24}}n^{4}+{\frac {1}{12}}nn}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <mstyle displaystyle="false" scriptlevel="0"> <mo>∫<!-- ∫ --></mo> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>7</mn> </mrow> </msup> <mo>=</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>8</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>8</mn> </mrow> </msup> <mo>+</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>7</mn> </mrow> </msup> <mo>+</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>7</mn> <mn>12</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>6</mn> </mrow> </msup> <mo>−<!-- − --></mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>7</mn> <mn>24</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>4</mn> </mrow> </msup> <mo>+</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>12</mn> </mfrac> </mrow> <mi>n</mi> <mi>n</mi> </mstyle> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle \textstyle \int n^{7}={\frac {1}{8}}n^{8}+{\frac {1}{2}}n^{7}+{\frac {7}{12}}n^{6}-{\frac {7}{24}}n^{4}+{\frac {1}{12}}nn}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/808a0fce328337a190c2ee4f075aa45105988461" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.338ex; width:42.056ex; height:3.843ex;" alt="{\displaystyle \textstyle \int n^{7}={\frac {1}{8}}n^{8}+{\frac {1}{2}}n^{7}+{\frac {7}{12}}n^{6}-{\frac {7}{24}}n^{4}+{\frac {1}{12}}nn}"/></span><br/><span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt124" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"\\textstyle \\int n^{8}=\\frac{1}{9}n^{9}+\\frac{1}{2}n^{8}+\\frac{2}{3}n^{7}-\\frac{7}{15}n^{5}+\\frac{2}{9}n^{3}-\\frac{1}{30}n"}}'><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle \textstyle \int n^{8}={\frac {1}{9}}n^{9}+{\frac {1}{2}}n^{8}+{\frac {2}{3}}n^{7}-{\frac {7}{15}}n^{5}+{\frac {2}{9}}n^{3}-{\frac {1}{30}}n}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <mstyle displaystyle="false" scriptlevel="0"> <mo>∫<!-- ∫ --></mo> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>8</mn> </mrow> </msup> <mo>=</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>9</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>9</mn> </mrow> </msup> <mo>+</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>8</mn> </mrow> </msup> <mo>+</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>2</mn> <mn>3</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>7</mn> </mrow> </msup> <mo>−<!-- − --></mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>7</mn> <mn>15</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>5</mn> </mrow> </msup> <mo>+</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>2</mn> <mn>9</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>3</mn> </mrow> </msup> <mo>−<!-- − --></mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>30</mn> </mfrac> </mrow> <mi>n</mi> </mstyle> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle \textstyle \int n^{8}={\frac {1}{9}}n^{9}+{\frac {1}{2}}n^{8}+{\frac {2}{3}}n^{7}-{\frac {7}{15}}n^{5}+{\frac {2}{9}}n^{3}-{\frac {1}{30}}n}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/c5a7fd18a7f196ada3761937e7f8a6e40dbc848a" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.338ex; width:46.787ex; height:3.843ex;" alt="{\displaystyle \textstyle \int n^{8}={\frac {1}{9}}n^{9}+{\frac {1}{2}}n^{8}+{\frac {2}{3}}n^{7}-{\frac {7}{15}}n^{5}+{\frac {2}{9}}n^{3}-{\frac {1}{30}}n}"/></span><br/><span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt125" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"\\textstyle \\int n^{9}=\\frac{1}{10}n^{10}+\\frac{1}{2}n^{9}+\\frac{3}{4}n^{8}-\\frac{7}{10}n^{6}+\\frac{1}{2}n^{4}-\\frac{1}{12}nn"}}'><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle \textstyle \int n^{9}={\frac {1}{10}}n^{10}+{\frac {1}{2}}n^{9}+{\frac {3}{4}}n^{8}-{\frac {7}{10}}n^{6}+{\frac {1}{2}}n^{4}-{\frac {1}{12}}nn}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <mstyle displaystyle="false" scriptlevel="0"> <mo>∫<!-- ∫ --></mo> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>9</mn> </mrow> </msup> <mo>=</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>10</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>10</mn> </mrow> </msup> <mo>+</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>9</mn> </mrow> </msup> <mo>+</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>3</mn> <mn>4</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>8</mn> </mrow> </msup> <mo>−<!-- − --></mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>7</mn> <mn>10</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>6</mn> </mrow> </msup> <mo>+</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>4</mn> </mrow> </msup> <mo>−<!-- − --></mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>12</mn> </mfrac> </mrow> <mi>n</mi> <mi>n</mi> </mstyle> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle \textstyle \int n^{9}={\frac {1}{10}}n^{10}+{\frac {1}{2}}n^{9}+{\frac {3}{4}}n^{8}-{\frac {7}{10}}n^{6}+{\frac {1}{2}}n^{4}-{\frac {1}{12}}nn}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/40a58ae34a7b88205f3f492f7d6f73742582ec29" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.338ex; width:49.825ex; height:3.843ex;" alt="{\displaystyle \textstyle \int n^{9}={\frac {1}{10}}n^{10}+{\frac {1}{2}}n^{9}+{\frac {3}{4}}n^{8}-{\frac {7}{10}}n^{6}+{\frac {1}{2}}n^{4}-{\frac {1}{12}}nn}"/></span><br/><span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt126" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"\\textstyle \\int n^{10}=\\frac{1}{11}n^{11}+\\frac{1}{2}n^{10}+\\frac{5}{6}n^{9}-1n^{7}+1n^{5}-\\frac{1}{2}n^{3}+\\frac{5}{66}n"}}'><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle \textstyle \int n^{10}={\frac {1}{11}}n^{11}+{\frac {1}{2}}n^{10}+{\frac {5}{6}}n^{9}-1n^{7}+1n^{5}-{\frac {1}{2}}n^{3}+{\frac {5}{66}}n}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <mstyle displaystyle="false" scriptlevel="0"> <mo>∫<!-- ∫ --></mo> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>10</mn> </mrow> </msup> <mo>=</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>11</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>11</mn> </mrow> </msup> <mo>+</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>10</mn> </mrow> </msup> <mo>+</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>5</mn> <mn>6</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>9</mn> </mrow> </msup> <mo>−<!-- − --></mo> <mn>1</mn> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>7</mn> </mrow> </msup> <mo>+</mo> <mn>1</mn> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>5</mn> </mrow> </msup> <mo>−<!-- − --></mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>3</mn> </mrow> </msup> <mo>+</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>5</mn> <mn>66</mn> </mfrac> </mrow> <mi>n</mi> </mstyle> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle \textstyle \int n^{10}={\frac {1}{11}}n^{11}+{\frac {1}{2}}n^{10}+{\frac {5}{6}}n^{9}-1n^{7}+1n^{5}-{\frac {1}{2}}n^{3}+{\frac {5}{66}}n}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/50559607471e6db9639abfbf45f103b6615fb166" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.338ex; width:55.209ex; height:3.676ex;" alt="{\displaystyle \textstyle \int n^{10}={\frac {1}{11}}n^{11}+{\frac {1}{2}}n^{10}+{\frac {5}{6}}n^{9}-1n^{7}+1n^{5}-{\frac {1}{2}}n^{3}+{\frac {5}{66}}n}"/></span><br/>Quin imò qui legem progressionis inibi attentius inspexerit, eundem etiam continuare poterit absque his ratiociniorum ambagibus: Sumtâ enim <span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt127" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"c"}}'><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle c}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <mi>c</mi> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle c}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/86a67b81c2de995bd608d5b2df50cd8cd7d92455" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -0.338ex; width:1.007ex; height:1.676ex;" alt="{\displaystyle c}"/></span> pro potestatis cujuslibet exponente, fit summa omnium <span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt128" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"n^{c}"}}'><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle n^{c}}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mi>c</mi> </mrow> </msup> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle n^{c}}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/3a62b39da47c098d4cec1ac794318335b381c436" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -0.338ex; width:2.339ex; height:2.343ex;" alt="{\displaystyle n^{c}}"/></span> seu<br/><span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt129" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"\\textstyle \\int n^{c}= \\frac{1}{c+1}n^{c+1}+\\frac{1}{2}n^{c}+\\frac{c}{2}An^{c-1}+\\frac{c\\cdot c-1\\cdot c-2}{2\\cdot3\\cdot4}Bn^{c-3}"}}'><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle \textstyle \int n^{c}={\frac {1}{c+1}}n^{c+1}+{\frac {1}{2}}n^{c}+{\frac {c}{2}}An^{c-1}+{\frac {c\cdot c-1\cdot c-2}{2\cdot 3\cdot 4}}Bn^{c-3}}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <mstyle displaystyle="false" scriptlevel="0"> <mo>∫<!-- ∫ --></mo> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mi>c</mi> </mrow> </msup> <mo>=</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mrow> <mi>c</mi> <mo>+</mo> <mn>1</mn> </mrow> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mi>c</mi> <mo>+</mo> <mn>1</mn> </mrow> </msup> <mo>+</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> </mrow> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mi>c</mi> </mrow> </msup> <mo>+</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mi>c</mi> <mn>2</mn> </mfrac> </mrow> <mi>A</mi> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mi>c</mi> <mo>−<!-- − --></mo> <mn>1</mn> </mrow> </msup> <mo>+</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mrow> <mi>c</mi> <mo>⋅<!-- ⋅ --></mo> <mi>c</mi> <mo>−<!-- − --></mo> <mn>1</mn> <mo>⋅<!-- ⋅ --></mo> <mi>c</mi> <mo>−<!-- − --></mo> <mn>2</mn> </mrow> <mrow> <mn>2</mn> <mo>⋅<!-- ⋅ --></mo> <mn>3</mn> <mo>⋅<!-- ⋅ --></mo> <mn>4</mn> </mrow> </mfrac> </mrow> <mi>B</mi> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mi>c</mi> <mo>−<!-- − --></mo> <mn>3</mn> </mrow> </msup> </mstyle> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle \textstyle \int n^{c}={\frac {1}{c+1}}n^{c+1}+{\frac {1}{2}}n^{c}+{\frac {c}{2}}An^{c-1}+{\frac {c\cdot c-1\cdot c-2}{2\cdot 3\cdot 4}}Bn^{c-3}}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/0db038bd019e7bba87eae825201c954c6e91ee56" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.338ex; width:49.981ex; height:3.843ex;" alt="{\displaystyle \textstyle \int n^{c}={\frac {1}{c+1}}n^{c+1}+{\frac {1}{2}}n^{c}+{\frac {c}{2}}An^{c-1}+{\frac {c\cdot c-1\cdot c-2}{2\cdot 3\cdot 4}}Bn^{c-3}}"/></span><br/><span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt130" data-mw='{"name":"math","attrs":{},"body":{"extsrc":" \\qquad +\\frac{c\\cdot c-1\\cdot c-2\\cdot c-3\\cdot c-4}{2\\cdot3\\cdot4\\cdot5\\cdot6}Cn^{c-5}"}}'><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle \qquad +{\frac {c\cdot c-1\cdot c-2\cdot c-3\cdot c-4}{2\cdot 3\cdot 4\cdot 5\cdot 6}}Cn^{c-5}}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <mspace width="2em"></mspace> <mo>+</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mrow> <mi>c</mi> <mo>⋅<!-- ⋅ --></mo> <mi>c</mi> <mo>−<!-- − --></mo> <mn>1</mn> <mo>⋅<!-- ⋅ --></mo> <mi>c</mi> <mo>−<!-- − --></mo> <mn>2</mn> <mo>⋅<!-- ⋅ --></mo> <mi>c</mi> <mo>−<!-- − --></mo> <mn>3</mn> <mo>⋅<!-- ⋅ --></mo> <mi>c</mi> <mo>−<!-- − --></mo> <mn>4</mn> </mrow> <mrow> <mn>2</mn> <mo>⋅<!-- ⋅ --></mo> <mn>3</mn> <mo>⋅<!-- ⋅ --></mo> <mn>4</mn> <mo>⋅<!-- ⋅ --></mo> <mn>5</mn> <mo>⋅<!-- ⋅ --></mo> <mn>6</mn> </mrow> </mfrac> </mrow> <mi>C</mi> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mi>c</mi> <mo>−<!-- − --></mo> <mn>5</mn> </mrow> </msup> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle \qquad +{\frac {c\cdot c-1\cdot c-2\cdot c-3\cdot c-4}{2\cdot 3\cdot 4\cdot 5\cdot 6}}Cn^{c-5}}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/510f63ae8f11de81ab9637cb352348a5abb35fa9" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -2.005ex; width:41.257ex; height:5.343ex;" alt="{\displaystyle \qquad +{\frac {c\cdot c-1\cdot c-2\cdot c-3\cdot c-4}{2\cdot 3\cdot 4\cdot 5\cdot 6}}Cn^{c-5}}"/></span><br/><span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt131" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"\\qquad +\\frac{c\\cdot c-1\\cdot c-2\\cdot c-3\\cdot c-4\\cdot c-5\\cdot c-6}{2\\cdot3\\cdot4\\cdot5\\cdot6\\cdot7\\cdot8}Dn^{c-7}\\ldots"}}'><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle \qquad +{\frac {c\cdot c-1\cdot c-2\cdot c-3\cdot c-4\cdot c-5\cdot c-6}{2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8}}Dn^{c-7}\ldots }"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <mspace width="2em"></mspace> <mo>+</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mrow> <mi>c</mi> <mo>⋅<!-- ⋅ --></mo> <mi>c</mi> <mo>−<!-- − --></mo> <mn>1</mn> <mo>⋅<!-- ⋅ --></mo> <mi>c</mi> <mo>−<!-- − --></mo> <mn>2</mn> <mo>⋅<!-- ⋅ --></mo> <mi>c</mi> <mo>−<!-- − --></mo> <mn>3</mn> <mo>⋅<!-- ⋅ --></mo> <mi>c</mi> <mo>−<!-- − --></mo> <mn>4</mn> <mo>⋅<!-- ⋅ --></mo> <mi>c</mi> <mo>−<!-- − --></mo> <mn>5</mn> <mo>⋅<!-- ⋅ --></mo> <mi>c</mi> <mo>−<!-- − --></mo> <mn>6</mn> </mrow> <mrow> <mn>2</mn> <mo>⋅<!-- ⋅ --></mo> <mn>3</mn> <mo>⋅<!-- ⋅ --></mo> <mn>4</mn> <mo>⋅<!-- ⋅ --></mo> <mn>5</mn> <mo>⋅<!-- ⋅ --></mo> <mn>6</mn> <mo>⋅<!-- ⋅ --></mo> <mn>7</mn> <mo>⋅<!-- ⋅ --></mo> <mn>8</mn> </mrow> </mfrac> </mrow> <mi>D</mi> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mi>c</mi> <mo>−<!-- − --></mo> <mn>7</mn> </mrow> </msup> <mo>…<!-- … --></mo> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle \qquad +{\frac {c\cdot c-1\cdot c-2\cdot c-3\cdot c-4\cdot c-5\cdot c-6}{2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8}}Dn^{c-7}\ldots }</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/cfa7f7a254c816086569f71431754c2ea73c7877" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -2.005ex; width:57.903ex; height:5.343ex;" alt="{\displaystyle \qquad +{\frac {c\cdot c-1\cdot c-2\cdot c-3\cdot c-4\cdot c-5\cdot c-6}{2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8}}Dn^{c-7}\ldots }"/></span><br/>& ita deinceps, exponentem potestatis ipsius <span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt132" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"n"}}'><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle n}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <mi>n</mi> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle n}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/a601995d55609f2d9f5e233e36fbe9ea26011b3b" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -0.338ex; width:1.395ex; height:1.676ex;" alt="{\displaystyle n}"/></span> continuè minuendo binario, quosque perveniatur ad <span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt133" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"n"}}'><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle n}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <mi>n</mi> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle n}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/a601995d55609f2d9f5e233e36fbe9ea26011b3b" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -0.338ex; width:1.395ex; height:1.676ex;" alt="{\displaystyle n}"/></span> vel <span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt134" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"nn"}}'><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle nn}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <mi>n</mi> <mi>n</mi> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle nn}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/799fbff5c0aa6585bde445dbd82befc6df9a2e04" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -0.338ex; width:2.789ex; height:1.676ex;" alt="{\displaystyle nn}"/></span>. Literæ capitales <span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt135" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"A"}}'><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle A}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <mi>A</mi> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle A}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/7daff47fa58cdfd29dc333def748ff5fa4c923e3" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -0.338ex; width:1.743ex; height:2.176ex;" alt="{\displaystyle A}"/></span>, <span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt136" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"B"}}'><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle B}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <mi>B</mi> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle B}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/47136aad860d145f75f3eed3022df827cee94d7a" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -0.338ex; width:1.764ex; height:2.176ex;" alt="{\displaystyle B}"/></span>, <span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt137" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"C"}}'><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle C}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <mi>C</mi> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle C}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/4fc55753007cd3c18576f7933f6f089196732029" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -0.338ex; width:1.766ex; height:2.176ex;" alt="{\displaystyle C}"/></span>, <span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt138" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"D"}}'><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle D}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <mi>D</mi> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle D}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/f34a0c600395e5d4345287e21fb26efd386990e6" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -0.338ex; width:1.924ex; height:2.176ex;" alt="{\displaystyle D}"/></span> &c. ordine denotant coëfficientes ultimorum terminorum pro <span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt139" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"\\textstyle \\int nn"}}'><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle \textstyle \int nn}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <mstyle displaystyle="false" scriptlevel="0"> <mo>∫<!-- ∫ --></mo> <mi>n</mi> <mi>n</mi> </mstyle> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle \textstyle \int nn}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/39fb66574be33c1d0b13e6af10f2ce208f3d98f1" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.005ex; width:4.594ex; height:3.176ex;" alt="{\displaystyle \textstyle \int nn}"/></span>, <span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt140" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"\\textstyle \\int n^{4}"}}'><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle \textstyle \int n^{4}}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <mstyle displaystyle="false" scriptlevel="0"> <mo>∫<!-- ∫ --></mo> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>4</mn> </mrow> </msup> </mstyle> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle \textstyle \int n^{4}}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/d91cf8669e30d850fb276d136c969612b351dd48" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.005ex; width:4.254ex; height:3.176ex;" alt="{\displaystyle \textstyle \int n^{4}}"/></span>, <span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt141" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"\\textstyle \\int n^{6}"}}'><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle \textstyle \int n^{6}}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <mstyle displaystyle="false" scriptlevel="0"> <mo>∫<!-- ∫ --></mo> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>6</mn> </mrow> </msup> </mstyle> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle \textstyle \int n^{6}}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/304aa7c3926b1572c11f730563718e5f09010ff5" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.005ex; width:4.254ex; height:3.176ex;" alt="{\displaystyle \textstyle \int n^{6}}"/></span>, <span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt142" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"\\textstyle \\int n^{8}"}}'><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle \textstyle \int n^{8}}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <mstyle displaystyle="false" scriptlevel="0"> <mo>∫<!-- ∫ --></mo> <msup> <mi>n</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>8</mn> </mrow> </msup> </mstyle> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle \textstyle \int n^{8}}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/8fc9ddbe581ce8801cdc7f7904cc273c91c21f14" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.005ex; width:4.254ex; height:3.176ex;" alt="{\displaystyle \textstyle \int n^{8}}"/></span> &c. nempe<br/><span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt143" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"\\textstyle A=\\frac{1}{6},B=-\\frac{1}{30},C=\\frac{1}{42},D=-\\frac{1}{30}."}}'><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle \textstyle A={\frac {1}{6}},B=-{\frac {1}{30}},C={\frac {1}{42}},D=-{\frac {1}{30}}.}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <mstyle displaystyle="false" scriptlevel="0"> <mi>A</mi> <mo>=</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>6</mn> </mfrac> </mrow> <mo>,</mo> <mi>B</mi> <mo>=</mo> <mo>−<!-- − --></mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>30</mn> </mfrac> </mrow> <mo>,</mo> <mi>C</mi> <mo>=</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>42</mn> </mfrac> </mrow> <mo>,</mo> <mi>D</mi> <mo>=</mo> <mo>−<!-- − --></mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>30</mn> </mfrac> </mrow> <mo>.</mo> </mstyle> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle \textstyle A={\frac {1}{6}},B=-{\frac {1}{30}},C={\frac {1}{42}},D=-{\frac {1}{30}}.}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/b0784146f99232c4a9915d58e40c78814b3427ba" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.338ex; width:36.055ex; height:3.676ex;" alt="{\displaystyle \textstyle A={\frac {1}{6}},B=-{\frac {1}{30}},C={\frac {1}{42}},D=-{\frac {1}{30}}.}"/></span> <br/><br/> From <b>Jacob Bernoulli's <a rel="mw:WikiLink" href="./Ars_Conjectandi" title="Ars Conjectandi">Ars Conjectandi</a>, 1713</b></small><sup about="#mwt160" class="mw-ref reference" id="cite_ref-13" rel="dc:references" typeof="mw:Extension/ref" data-mw='{"name":"ref","attrs":{"group":"lower-alpha"},"body":{"id":"mw-reference-text-cite_note-13"}}'><a href="./Bernoulli_number#cite_note-13" data-mw-group="lower-alpha" id="mwAUE"><span class="mw-reflink-text" id="mwAUI"><span class="cite-bracket" id="mwAUM">[</span>a<span class="cite-bracket" id="mwAUQ">]</span></span></a></sup><sup about="#mwt175" class="mw-ref reference" id="cite_ref-Bernoulli1713_14-0" rel="dc:references" typeof="mw:Extension/ref" data-mw='{"name":"ref","attrs":{"name":"Bernoulli1713"}}'><a href="./Bernoulli_number#cite_note-Bernoulli1713-14" id="mwAUU"><span class="mw-reflink-text" id="mwAUY"><span class="cite-bracket" id="mwAUc">[</span>13<span class="cite-bracket" id="mwAUg">]</span></span></a></sup>

The Bernoulli numbers OEIS: A164555(n)/OEIS: A027642(n) were introduced by Jacob Bernoulli in the book Ars Conjectandi published posthumously in 1713. The main formula can be seen in the second half of the corresponding facsimile. The constant coefficients denoted A, B, C and D by Bernoulli are mapped to the notation which is now prevalent as A = B₂, B = B₄, C = B₆, D = B₈. The expression c·c−1·c−2·c−3 means c·(c−1)·(c−2)·(c−3) – the small dots are used as grouping symbols. Using today's terminology these expressions are falling factorial powers c^k. The factorial notation k! as a shortcut for 1 × 2 × ... × k was introduced much later in 1808 by Christian Kramp. The integral symbol ∫ on the left hand side goes back to Gottfried Wilhelm Leibniz in 1675 who used it as a long letter S for "summa" (sum). The letter n on the left hand side is not an index of summation but gives the upper limit of the range of summation which is to be understood as 1, 2, ..., n. Putting things together, for positive c, today a mathematician is likely to write Bernoulli's formula as:

\sum_{k = 1}^{n} k^{c} = \frac{n^{c + 1}}{c + 1} + \frac{1}{2} n^{c} + \sum_{k = 2}^{c} \frac{B_{k}}{k!} c^{\underline{k - 1}} n^{c - k + 1} .

This formula suggests setting B₁ = ⁠1/2⁠ when switching from the so-called 'archaic' enumeration which uses only the even indices 2, 4, 6... to the modern form. Most striking in this context is the fact that the falling factorial c^k−1 has for k = 0 the value ⁠1/c + 1⁠. Thus Bernoulli's formula can be written

\sum_{k = 1}^{n} k^{c} = \sum_{k = 0}^{c} \frac{B_{k}}{k!} c^{\underline{k - 1}} n^{c - k + 1}

if B₁ = 1/2, recapturing the value Bernoulli gave to the coefficient at that position.

The formula for $\sum_{k = 1}^{n} k^{9}$ on page 97 of Bernoulli's Ars Conjectandi contains an error at the last term; it should be $- \frac{3}{20} n^{2}$ instead of $- \frac{1}{12} n^{2}$ .

Definitions

Many characterizations of the Bernoulli numbers have been found in the last 300 years, and each could be used to introduce these numbers. Here only four of the most useful ones are mentioned:

a recursive equation,
an explicit formula,
a generating function,
an integral expression.

For the proof of the equivalence of the four approaches, see Ireland & Rosen (1990) or Conway & Guy (1996).

Recursive definition

The Bernoulli numbers obey the sum formulas

\begin{matrix} \sum_{k = 0}^{m} (\binom{m + 1}{k}) B_{k}^{-} & = δ_{m, 0} \\ \sum_{k = 0}^{m} (\binom{m + 1}{k}) B_{k}^{+} & = m + 1 \end{matrix}

where $m = 0, 1, 2...$ and δ denotes the Kronecker delta.

The first of these is sometimes written as the formula (for m > 1) $(B + 1)^{m} - B_{m} = 0,$ where the power is expanded formally using the binomial theorem and $B^{k}$ is replaced by $B_{k}$ .

Solving for $B_{m}^{\mp}$ gives the recursive formulas

\begin{matrix} B_{m}^{-} & = δ_{m, 0} - \sum_{k = 0}^{m - 1} (\binom{m}{k}) \frac{B_{k}^{-}}{m - k + 1} \\ B_{m}^{+} & = 1 - \sum_{k = 0}^{m - 1} (\binom{m}{k}) \frac{B_{k}^{+}}{m - k + 1} . \end{matrix}

Explicit definition

In 1893 Louis Saalschütz listed a total of 38 explicit formulas for the Bernoulli numbers, usually giving some reference in the older literature. One of them is (for $m \geq 1$ ):

\begin{matrix} B_{m}^{-} & = \sum_{k = 0}^{m} \frac{1}{k + 1} \sum_{j = 0}^{k} (\binom{k}{j}) (- 1)^{j} j^{m} \\ B_{m}^{+} & = \sum_{k = 0}^{m} \frac{1}{k + 1} \sum_{j = 0}^{k} (\binom{k}{j}) (- 1)^{j} (j + 1)^{m} . \end{matrix}

Generating function

The exponential generating functions are

\begin{aligned} \frac{t}{e^{t} - 1} & = \frac{t}{2} (\coth \frac{t}{2} - 1) & = \sum_{m = 0}^{\infty} \frac{B_{m}^{-} t^{m}}{m!} \\ \frac{t e^{t}}{e^{t} - 1} = \frac{t}{1 - e^{- t}} & = \frac{t}{2} (\coth \frac{t}{2} + 1) & = \sum_{m = 0}^{\infty} \frac{B_{m}^{+} t^{m}}{m!} . \end{aligned}

where the substitution is $t \to - t$ . The arithmetic difference between the generating functions for $B_{m}^{+}$ and $B_{m}^{-}$ is t.

The (ordinary) generating function

z^{- 1} ψ_{1} (z^{- 1}) = \sum_{m = 0}^{\infty} B_{m}^{+} z^{m}

is an asymptotic series. It contains the trigamma function ψ₁.

Integral Expression

From the generating functions above, one can obtain the following integral formula for the even Bernoulli numbers:

B_{2 n} = 4 n (- 1)^{n + 1} \int_{0}^{\infty} \frac{t^{2 n - 1}}{e^{2 π t} - 1} d t

Bernoulli numbers and the Riemann zeta function

The Bernoulli numbers can be expressed in terms of the Riemann zeta function:

B_{n}^{+} = - n ζ (1 - n)

for n ≥ 1 .

Here the argument of the zeta function is 0 or negative. As $ζ (k)$ is zero for negative even integers (the trivial zeroes), if n>1 is odd, $ζ (1 - n)$ is zero.

By means of the zeta functional equation and the gamma reflection formula the following relation can be obtained:

B_{2 n} = \frac{(- 1)^{n + 1} 2 (2 n)!}{(2 π)^{2 n}} ζ (2 n)

for integers n ≥ 1 .

Now the argument of the zeta function is positive.

It then follows from ζ → 1 (n → ∞) and Stirling's formula that

| B_{2 n} | \sim 4 \sqrt{π n} {(\frac{n}{π e})}^{2 n}

for n → ∞ .

Efficient computation of Bernoulli numbers

Akiyama and Tanigawa give a simple "triangle algorithm" (vis-à-vis Pascal's triangle) for computing the Bernoulli numbers. First let $b_{0, m} = \frac{1}{m + 1}$ (for m ≥ 0). Then successive terms in the triangle can be computed with the recurrence relation

$b_{n + 1, m} = (m + 1) (b_{n, m} - b_{n, m + 1})$

The terms $b_{n, 0}$ correspond to the nth Bernoulli number B_n.

Brent and Harvey give several algorithms for computing the Bernoulli numbers, including a simple algorithm that is faster and uses less space than the Akiyama-Tanigawa algorithm. It uses a recurrence to compute the tangent numbers T_n and applies

$T_{n} = (- 1)^{n - 1} 2^{2 n} (2^{2 n} - 1) \frac{B_{2 n}}{2 n}$

to compute the Bernoulli numbers. This is related to a method given by Knuth and Buckholtz.

In some applications it is useful to be able to compute the Bernoulli numbers B₀ through B_{p − 3} modulo p, where p is a prime; for example to test whether Vandiver's conjecture holds for p, or even just to determine whether p is an irregular prime. It is not feasible to carry out such a computation using the above recursive formulae, since at least (a constant multiple of) p² arithmetic operations would be required. Fortunately, faster methods have been developed which require only O(p (log p)²) operations (see big O notation).

David Harvey describes an algorithm for computing Bernoulli numbers by computing B_n modulo p for many small primes p, and then reconstructing B_n via the Chinese remainder theorem. Harvey writes that the asymptotic time complexity of this algorithm is O(n² log(n)^{2 + ε}) and claims that this implementation is significantly faster than implementations based on other methods. Using this implementation Harvey computed B_n for n = 10⁸. Harvey's implementation has been included in SageMath since version 3.1. Prior to that, Bernd Kellner computed B_n to full precision for n = 10⁶ in December 2002 and Oleksandr Pavlyk for n = 10⁷ with Mathematica in April 2008.

* Digits is to be understood as the exponent of 10 when B_n is written as a real number in normalized scientific notation.

Applications of the Bernoulli numbers

Asymptotic analysis

Arguably the most important application of the Bernoulli numbers in mathematics is their use in the Euler–Maclaurin formula. Assuming that f is a sufficiently often differentiable function the Euler–Maclaurin formula can be written as

\sum_{k = a}^{b - 1} f (k) = \int_{a}^{b} f (x) d x + \sum_{k = 1}^{m} \frac{B_{k}^{-}}{k!} (f^{(k - 1)} (b) - f^{(k - 1)} (a)) + R_{-} (f, m) .

This formulation assumes the convention B⁻
₁ = −⁠1/2⁠. Using the convention B⁺
₁ = +⁠1/2⁠ the formula becomes

\sum_{k = a + 1}^{b} f (k) = \int_{a}^{b} f (x) d x + \sum_{k = 1}^{m} \frac{B_{k}^{+}}{k!} (f^{(k - 1)} (b) - f^{(k - 1)} (a)) + R_{+} (f, m) .

Here $f^{(0)} = f$ (i.e. the zeroth-order derivative of $f$ is just $f$ ). Moreover, let $f^{(- 1)}$ denote an antiderivative of $f$ . By the fundamental theorem of calculus,

\int_{a}^{b} f (x) d x = f^{(- 1)} (b) - f^{(- 1)} (a) .

Thus the last formula can be further simplified to the following succinct form of the Euler–Maclaurin formula

\sum_{k = a + 1}^{b} f (k) = \sum_{k = 0}^{m} \frac{B_{k}}{k!} (f^{(k - 1)} (b) - f^{(k - 1)} (a)) + R (f, m) .

This form is for example the source for the important Euler–Maclaurin expansion of the zeta function

\begin{matrix} ζ (s) & = \sum_{k = 0}^{m} \frac{B_{k}^{+}}{k!} s^{\bar{k - 1}} + R (s, m) \\ = \frac{B_{0}}{0!} s^{\bar{- 1}} + \frac{B_{1}^{+}}{1!} s^{\bar{0}} + \frac{B_{2}}{2!} s^{\bar{1}} + \dots + R (s, m) \\ = \frac{1}{s - 1} + \frac{1}{2} + \frac{1}{12} s + \dots + R (s, m) . \end{matrix}

Here s^k denotes the rising factorial power.

Bernoulli numbers are also frequently used in other kinds of asymptotic expansions. The following example is the classical Poincaré-type asymptotic expansion of the digamma function ψ.

ψ (z) \sim \ln z - \sum_{k = 1}^{\infty} \frac{B_{k}^{+}}{k z^{k}}

Sum of powers

Bernoulli numbers feature prominently in the closed form expression of the sum of the mth powers of the first n positive integers. For m, n ≥ 0 define

S_{m} (n) = \sum_{k = 1}^{n} k^{m} = 1^{m} + 2^{m} + \dots + n^{m} .

This expression can always be rewritten as a polynomial in n of degree m + 1. The coefficients of these polynomials are related to the Bernoulli numbers by Bernoulli's formula:

S_{m} (n) = \frac{1}{m + 1} \sum_{k = 0}^{m} (\binom{m + 1}{k}) B_{k}^{+} n^{m + 1 - k} = m! \sum_{k = 0}^{m} \frac{B_{k}^{+} n^{m + 1 - k}}{k! (m + 1 - k)!},

where (^{m + 1}
_k) denotes the binomial coefficient.

For example, taking m to be 1 gives the triangular numbers 0, 1, 3, 6, ... OEIS: A000217.

1 + 2 + \dots + n = \frac{1}{2} (B_{0} n^{2} + 2 B_{1}^{+} n^{1}) = \frac{1}{2} (n^{2} + n) .

Taking m to be 2 gives the square pyramidal numbers 0, 1, 5, 14, ... OEIS: A000330.

1^{2} + 2^{2} + \dots + n^{2} = \frac{1}{3} (B_{0} n^{3} + 3 B_{1}^{+} n^{2} + 3 B_{2} n^{1}) = \frac{1}{3} (n^{3} + \frac{3}{2} n^{2} + \frac{1}{2} n) .

Some authors use the alternate convention for Bernoulli numbers and state Bernoulli's formula in this way:

S_{m} (n) = \frac{1}{m + 1} \sum_{k = 0}^{m} (- 1)^{k} (\binom{m + 1}{k}) B_{k}^{-} n^{m + 1 - k} .

Bernoulli's formula is sometimes called Faulhaber's formula after Johann Faulhaber who also found remarkable ways to calculate sums of powers.

Faulhaber's formula was generalized by V. Guo and J. Zeng to a q-analog.

Taylor series

The Bernoulli numbers appear in the Taylor series expansion of many trigonometric functions and hyperbolic functions.

$\begin{matrix} \tan x & = \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} 2^{2 n} (2^{2 n} - 1) B_{2 n}}{(2 n)!} x^{2 n - 1}, & | x | < \frac{π}{2} . \\ \cot x & = \frac{1}{x} \sum_{n = 0}^{\infty} \frac{(- 1)^{n} B_{2 n} (2 x)^{2 n}}{(2 n)!}, & 0 < & | x | < π . \\ \tanh x & = \sum_{n = 1}^{\infty} \frac{2^{2 n} (2^{2 n} - 1) B_{2 n}}{(2 n)!} x^{2 n - 1}, & | x | < \frac{π}{2} . \\ \coth x & = \frac{1}{x} \sum_{n = 0}^{\infty} \frac{B_{2 n} (2 x)^{2 n}}{(2 n)!}, & 0 < & | x | < π . \end{matrix}$

Laurent series

The Bernoulli numbers appear in the following Laurent series:

Digamma function: $ψ (z) = \ln z - \sum_{k = 1}^{\infty} \frac{B_{k}^{+}}{k z^{k}}$

Use in topology

The Kervaire–Milnor formula for the order of the cyclic group of diffeomorphism classes of exotic (4n − 1)-spheres which bound parallelizable manifolds involves Bernoulli numbers. Let ES_n be the number of such exotic spheres for n ≥ 2, then

{ES}_{n} = (2^{2 n - 2} - 2^{4 n - 3}) Numerator (\frac{B_{4 n}}{4 n}) .

The Hirzebruch signature theorem for the L genus of a smooth oriented closed manifold of dimension 4n also involves Bernoulli numbers.

Connections with combinatorial numbers

The connection of the Bernoulli number to various kinds of combinatorial numbers is based on the classical theory of finite differences and on the combinatorial interpretation of the Bernoulli numbers as an instance of a fundamental combinatorial principle, the inclusion–exclusion principle.

Connection with Worpitzky numbers

The definition to proceed with was developed by Julius Worpitzky in 1883. Besides elementary arithmetic only the factorial function n! and the power function k^m is employed. The signless Worpitzky numbers are defined as

W_{n, k} = \sum_{v = 0}^{k} (- 1)^{v + k} (v + 1)^{n} (\binom{k}{v}) .

They can also be expressed through the Stirling numbers of the second kind

W_{n, k} = k! {\binom{n + 1}{k + 1}} .

A Bernoulli number is then introduced as an inclusion–exclusion sum of Worpitzky numbers weighted by the harmonic sequence 1, ⁠1/2⁠, ⁠1/3⁠, ...

B_{n} = \sum_{k = 0}^{n} (- 1)^{k} \frac{W_{n, k}}{k + 1} = \sum_{k = 0}^{n} \frac{1}{k + 1} \sum_{v = 0}^{k} (- 1)^{v} (v + 1)^{n} (\binom{k}{v}) .

B₀ = 1

B₁ = 1 − ⁠1/2⁠

B₂ = 1 − ⁠3/2⁠ + ⁠2/3⁠

B₃ = 1 − ⁠7/2⁠ + ⁠12/3⁠ − ⁠6/4⁠

B₄ = 1 − ⁠15/2⁠ + ⁠50/3⁠ − ⁠60/4⁠ + ⁠24/5⁠

B₅ = 1 − ⁠31/2⁠ + ⁠180/3⁠ − ⁠390/4⁠ + ⁠360/5⁠ − ⁠120/6⁠

B₆ = 1 − ⁠63/2⁠ + ⁠602/3⁠ − ⁠2100/4⁠ + ⁠3360/5⁠ − ⁠2520/6⁠ + ⁠720/7⁠

This representation has B⁺
₁ = +⁠1/2⁠.

Consider the sequence s_n, n ≥ 0. From Worpitzky's numbers OEIS: A028246, OEIS: A163626 applied to s₀, s₀, s₁, s₀, s₁, s₂, s₀, s₁, s₂, s₃, ... is identical to the Akiyama–Tanigawa transform applied to s_n (see Connection with Stirling numbers of the first kind). This can be seen via the table:

The first row represents s₀, s₁, s₂, s₃, s₄.

Hence for the second fractional Euler numbers OEIS: A198631 (n) / OEIS: A006519 (n + 1):

E₀ = 1

E₁ = 1 − ⁠1/2⁠

E₂ = 1 − ⁠3/2⁠ + ⁠2/4⁠

E₃ = 1 − ⁠7/2⁠ + ⁠12/4⁠ − ⁠6/8⁠

E₄ = 1 − ⁠15/2⁠ + ⁠50/4⁠ − ⁠60/8⁠ + ⁠24/16⁠

E₅ = 1 − ⁠31/2⁠ + ⁠180/4⁠ − ⁠390/8⁠ + ⁠360/16⁠ − ⁠120/32⁠

E₆ = 1 − ⁠63/2⁠ + ⁠602/4⁠ − ⁠2100/8⁠ + ⁠3360/16⁠ − ⁠2520/32⁠ + ⁠720/64⁠

A second formula representing the Bernoulli numbers by the Worpitzky numbers is for n ≥ 1

B_{n} = \frac{n}{2^{n + 1} - 2} \sum_{k = 0}^{n - 1} (- 2)^{- k} W_{n - 1, k} .

The simplified second Worpitzky's representation of the second Bernoulli numbers is:

OEIS: A164555 (n + 1) / OEIS: A027642(n + 1) = ⁠n + 1/2^{n + 2} − 2⁠ × OEIS: A198631(n) / OEIS: A006519(n + 1)

which links the second Bernoulli numbers to the second fractional Euler numbers. The beginning is:

⁠1/2⁠, ⁠1/6⁠, 0, −⁠1/30⁠, 0, ⁠1/42⁠, ... = (⁠1/2⁠, ⁠1/3⁠, ⁠3/14⁠, ⁠2/15⁠, ⁠5/62⁠, ⁠1/21⁠, ...) × (1, ⁠1/2⁠, 0, −⁠1/4⁠, 0, ⁠1/2⁠, ...)

The numerators of the first parentheses are OEIS: A111701 (see Connection with Stirling numbers of the first kind).

Connection with Stirling numbers of the second kind

Stirling numbers of the second kind, S(k,m), have the property that:

x^{k} = \sum_{m = 0}^{k} x^{\underline{m}} S (k, m)

where x^m denotes the falling factorial function.

The Bernoulli polynomials B_k(x) can be written as:

B_{k} (x) = k \sum_{m = 0}^{k - 1} (\binom{x}{m + 1}) S (k - 1, m) m! + B_{k}

where B_k for k = 0, 1, 2,... are the Bernoulli numbers.

The following property of the binomial coefficient:

(\binom{x}{m}) = (\binom{x + 1}{m + 1}) - (\binom{x}{m + 1})

thus implies that

x^{k} = \frac{B_{k + 1} (x + 1) - B_{k + 1} (x)}{k + 1} .

One also has the following for Bernoulli polynomials,

B_{k} (x) = \sum_{n = 0}^{k} (\binom{k}{n}) B_{n} x^{k - n} .

The coefficient of x in (^x
_{m + 1}) is ⁠(−1)^m/m + 1⁠.

The coefficient of x in the first expression for $B_{k} (x)$ is $k \sum_{m = 0}^{k - 1} \frac{(- 1)^{m}}{m + 1} S (k - 1, m) m!$ whereas in the second expression it is $k B_{k - 1} .$ Replacing $k - 1$ by $k$ this yields:

B_{k} = \sum_{m = 0}^{k} (- 1)^{m} \frac{m!}{m + 1} S (k, m)

(resulting in B₁ = +⁠1/2⁠) which is an explicit formula for Bernoulli numbers and can be used to prove Von-Staudt Clausen theorem.

Connection with Stirling numbers of the first kind

The two main formulas relating the unsigned Stirling numbers of the first kind [ⁿ
_m] to the Bernoulli numbers (with B₁ = +⁠1/2⁠) are

\frac{1}{m!} \sum_{k = 0}^{m} (- 1)^{k} [\binom{m + 1}{k + 1}] B_{k} = \frac{1}{m + 1},

and the inversion of this sum (for n ≥ 0, m ≥ 0)

\frac{1}{m!} \sum_{k = 0}^{m} (- 1)^{k} [\binom{m + 1}{k + 1}] B_{n + k} = A_{n, m} .

Here the number A_n,m are the rational Akiyama–Tanigawa numbers, the first few of which are displayed in the following table.

The Akiyama–Tanigawa numbers satisfy a simple recurrence relation which can be exploited to iteratively compute the Bernoulli numbers. See OEIS: A051714/OEIS: A051715.

An autosequence is a sequence which has its inverse binomial transform equal to the signed sequence. If the main diagonal is zeroes = OEIS: A000004, the autosequence is of the first kind. Example: OEIS: A000045, the Fibonacci numbers. If the main diagonal is the first upper diagonal multiplied by 2, it is of the second kind. Example: OEIS: A164555/OEIS: A027642, the second Bernoulli numbers (see OEIS: A190339). The Akiyama–Tanigawa transform applied to 2⁻ⁿ = 1/OEIS: A000079 leads to OEIS: A198631 (n) / OEIS: A06519 (n + 1). Hence:

See OEIS: A209308 and OEIS: A227577. OEIS: A198631 (n) / OEIS: A006519 (n + 1) are the second (fractional) Euler numbers and an autosequence of the second kind.

(⁠OEIS: A164555 (n + 2)/OEIS: A027642 (n + 2)⁠ = ⁠1/6⁠, 0, −⁠1/30⁠, 0, ⁠1/42⁠, ...) × ( ⁠2^{n + 3} − 2/n + 2⁠ = 3, ⁠14/3⁠, ⁠15/2⁠, ⁠62/5⁠, 21, ...) = ⁠OEIS: A198631 (n + 1)/OEIS: A006519 (n + 2)⁠ = ⁠1/2⁠, 0, −⁠1/4⁠, 0, ⁠1/2⁠, ....

Also valuable for OEIS: A027641 / OEIS: A027642 (see Connection with Worpitzky numbers).

Connection with Pascal's triangle

There are formulas connecting Pascal's triangle to Bernoulli numbers

B_{n}^{+} = \frac{| A_{n} |}{(n + 1)!}

with row-column definitions

[A_{n}]_{i, k} := {\begin{cases} 0 & if k > 1 + i \\ (\binom{i + 1}{k - 1}) & otherwise \end{cases}

where $| A_{n} |$ is the determinant of a n-by-n Hessenberg matrix part of Pascal's triangle

Example:

B_{6}^{+} = \frac{| \begin{matrix} 1 & 2 & 0 & 0 & 0 & 0 \\ 1 & 3 & 3 & 0 & 0 & 0 \\ 1 & 4 & 6 & 4 & 0 & 0 \\ 1 & 5 & 10 & 10 & 5 & 0 \\ 1 & 6 & 15 & 20 & 15 & 6 \\ 1 & 7 & 21 & 35 & 35 & 21 \end{matrix} |}{7!} = \frac{120}{5040} = \frac{1}{42}

Connection with Eulerian numbers

There are formulas connecting Eulerian numbers ⟨ⁿ
_m⟩ to Bernoulli numbers:

\begin{matrix} \sum_{m = 0}^{n} (- 1)^{m} ⟨ \binom{n}{m} ⟩ & = 2^{n + 1} (2^{n + 1} - 1) \frac{B_{n + 1}}{n + 1}, \\ \sum_{m = 0}^{n} (- 1)^{m} ⟨ \binom{n}{m} ⟩ {(\binom{n}{m})}^{- 1} & = (n + 1) B_{n} . \end{matrix}

Both formulae are valid for n ≥ 0 if B₁ is set to ⁠1/2⁠. If B₁ is set to −⁠1/2⁠ they are valid only for n ≥ 1 and n ≥ 2 respectively.

A binary tree representation

The Stirling polynomials σ_n(x) are related to the Bernoulli numbers by B_n = n!σ_n(1). S. C. Woon described an algorithm to compute σ_n(1) as a binary tree:

Woon's recursive algorithm (for n ≥ 1) starts by assigning to the root node N = [1,2]. Given a node N = [a₁, a₂, ..., a_k] of the tree, the left child of the node is L(N) = [−a₁, a₂ + 1, a₃, ..., a_k] and the right child R(N) = [a₁, 2, a₂, ..., a_k]. A node N = [a₁, a₂, ..., a_k] is written as ±[a₂, ..., a_k] in the initial part of the tree represented above with ± denoting the sign of a₁.

Given a node N the factorial of N is defined as

N! = a_{1} \prod_{k = 2}^{length (N)} a_{k}! .

Restricted to the nodes N of a fixed tree-level n the sum of ⁠1/N!⁠ is σ_n(1), thus

B_{n} = \sum_{\overset{N node of}{tree-level n}} \frac{n!}{N!} .

For example:

B₁ = 1!(⁠1/2!⁠)

B₂ = 2!(−⁠1/3!⁠ + ⁠1/2!2!⁠)

B₃ = 3!(⁠1/4!⁠ − ⁠1/2!3!⁠ − ⁠1/3!2!⁠ + ⁠1/2!2!2!⁠)

Integral representation and continuation

The integral

b (s) = 2 e^{s i π / 2} \int_{0}^{\infty} \frac{s t^{s}}{1 - e^{2 π t}} \frac{d t}{t} = \frac{s!}{2^{s - 1}} \frac{ζ (s)}{π^{s}} (- i)^{s} = \frac{2 s! ζ (s)}{(2 π i)^{s}}

has as special values b(2n) = B_2n for n > 0.

For example, b(3) = ⁠3/2⁠ζ(3)π⁻³i and b(5) = −⁠15/2⁠ζ(5)π⁻⁵i. Here, ζ is the Riemann zeta function, and i is the imaginary unit. Leonhard Euler (Opera Omnia, Ser. 1, Vol. 10, p. 351) considered these numbers and calculated

\begin{matrix} p & = \frac{3}{2 π^{3}} (1 + \frac{1}{2^{3}} + \frac{1}{3^{3}} + \dots) = 0.0581522 \dots \\ q & = \frac{15}{2 π^{5}} (1 + \frac{1}{2^{5}} + \frac{1}{3^{5}} + \dots) = 0.0254132 \dots \end{matrix}

Another similar integral representation is

b (s) = - \frac{e^{s i π / 2}}{2^{s} - 1} \int_{0}^{\infty} \frac{s t^{s}}{\sinh π t} \frac{d t}{t} = \frac{2 e^{s i π / 2}}{2^{s} - 1} \int_{0}^{\infty} \frac{e^{π t} s t^{s}}{1 - e^{2 π t}} \frac{d t}{t} .

The relation to the Euler numbers and π

The Euler numbers are a sequence of integers intimately connected with the Bernoulli numbers. Comparing the asymptotic expansions of the Bernoulli and the Euler numbers shows that the Euler numbers E_2n are in magnitude approximately ⁠2/π⁠(4²ⁿ − 2²ⁿ) times larger than the Bernoulli numbers B_2n. In consequence:

π \sim 2 (2^{2 n} - 4^{2 n}) \frac{B_{2 n}}{E_{2 n}} .

This asymptotic equation reveals that π lies in the common root of both the Bernoulli and the Euler numbers. In fact π could be computed from these rational approximations.

Bernoulli numbers can be expressed through the Euler numbers and vice versa. Since, for odd n, B_n = E_n = 0 (with the exception B₁), it suffices to consider the case when n is even.

\begin{matrix} B_{n} & = \sum_{k = 0}^{n - 1} (\binom{n - 1}{k}) \frac{n}{4^{n} - 2^{n}} E_{k} & n & = 2, 4, 6, \dots \\ E_{n} & = \sum_{k = 1}^{n} (\binom{n}{k - 1}) \frac{2^{k} - 4^{k}}{k} B_{k} & n & = 2, 4, 6, \dots \end{matrix}

These conversion formulas express a connection between the Bernoulli and the Euler numbers. But more important, there is a deep arithmetic root common to both kinds of numbers, which can be expressed through a more fundamental sequence of numbers, also closely tied to π. These numbers are defined for n ≥ 1 as

S_{n} = 2 {(\frac{2}{π})}^{n} \sum_{k = 0}^{\infty} \frac{(- 1)^{k n}}{(2 k + 1)^{n}} = 2 {(\frac{2}{π})}^{n} lim_{K \to \infty} \sum_{k = - K}^{K} (4 k + 1)^{- n} .

The magic of these numbers lies in the fact that they turn out to be rational numbers. This was first proved by Leonhard Euler in a landmark paper De summis serierum reciprocarum (On the sums of series of reciprocals) and has fascinated mathematicians ever since. The first few of these numbers are

S_{n} = 1, 1, \frac{1}{2}, \frac{1}{3}, \frac{5}{24}, \frac{2}{15}, \frac{61}{720}, \frac{17}{315}, \frac{277}{8064}, \frac{62}{2835}, \dots

(OEIS: A099612 / OEIS: A099617)

These are the coefficients in the expansion of sec x + tan x.

The Bernoulli numbers and Euler numbers can be understood as special views of these numbers, selected from the sequence S_n and scaled for use in special applications.

\begin{matrix} B_{n} & = (- 1)^{⌊ \frac{n}{2} ⌋} [n even] \frac{n!}{2^{n} - 4^{n}} S_{n}, & n & = 2, 3, \dots \\ E_{n} & = (- 1)^{⌊ \frac{n}{2} ⌋} [n even] n! S_{n + 1} & n & = 0, 1, \dots \end{matrix}

The expression [n even] has the value 1 if n is even and 0 otherwise (Iverson bracket).

These identities show that the quotient of Bernoulli and Euler numbers at the beginning of this section is just the special case of R_n = ⁠2S_n/S_{n + 1}⁠ when n is even. The R_n are rational approximations to π and two successive terms always enclose the true value of π. Beginning with n = 1 the sequence starts (OEIS: A132049 / OEIS: A132050):

2, 4, 3, \frac{16}{5}, \frac{25}{8}, \frac{192}{61}, \frac{427}{136}, \frac{4352}{1385}, \frac{12465}{3968}, \frac{158720}{50521}, \dots ⟶ π .

These rational numbers also appear in the last paragraph of Euler's paper cited above.

Consider the Akiyama–Tanigawa transform for the sequence OEIS: A046978 (n + 2) / OEIS: A016116 (n + 1):

From the second, the numerators of the first column are the denominators of Euler's formula. The first column is −⁠1/2⁠ × OEIS: A163982.

An algorithmic view: the Seidel triangle

The sequence S_n has another unexpected yet important property: The denominators of S_n+1 divide the factorial n!. In other words: the numbers T_n = S_{n + 1} n!, sometimes called Euler zigzag numbers, are integers.

T_{n} = 1, 1, 1, 2, 5, 16, 61, 272, 1385, 7936, 50521, 353792, \dots n = 0, 1, 2, 3, \dots

(OEIS: A000111). See (OEIS: A253671).

Their exponential generating function is the sum of the secant and tangent functions.

\sum_{n = 0}^{\infty} T_{n} \frac{x^{n}}{n!} = \tan (\frac{π}{4} + \frac{x}{2}) = \sec x + \tan x

Thus the above representations of the Bernoulli and Euler numbers can be rewritten in terms of this sequence as

\begin{matrix} B_{n} & = (- 1)^{⌊ \frac{n}{2} ⌋} [n even] \frac{n}{2^{n} - 4^{n}} T_{n - 1} & n & \geq 2 \\ E_{n} & = (- 1)^{⌊ \frac{n}{2} ⌋} [n even] T_{n} & n & \geq 0 \end{matrix}

These identities make it easy to compute the Bernoulli and Euler numbers: the Euler numbers E_2n are given immediately by T_2n and the Bernoulli numbers B_2n are fractions obtained from T_{2n - 1} by some easy shifting, avoiding rational arithmetic.

What remains is to find a convenient way to compute the numbers T_n. However, already in 1877 Philipp Ludwig von Seidel published an ingenious algorithm, which makes it simple to calculate T_n.

Start by putting 1 in row 0 and let k denote the number of the row currently being filled
If k is odd, then put the number on the left end of the row k − 1 in the first position of the row k, and fill the row from the left to the right, with every entry being the sum of the number to the left and the number to the upper
At the end of the row duplicate the last number.
If k is even, proceed similar in the other direction.

Seidel's algorithm is in fact much more general (see the exposition of Dominique Dumont ) and was rediscovered several times thereafter.

Similar to Seidel's approach D. E. Knuth and T. J. Buckholtz gave a recurrence equation for the numbers T_2n and recommended this method for computing B_2n and E_2n 'on electronic computers using only simple operations on integers'.

V. I. Arnold rediscovered Seidel's algorithm and later Millar, Sloane and Young popularized Seidel's algorithm under the name boustrophedon transform.

Triangular form:

Only OEIS: A000657, with one 1, and OEIS: A214267, with two 1s, are in the OEIS.

Distribution with a supplementary 1 and one 0 in the following rows:

This is OEIS: A239005, a signed version of OEIS: A008280. The main andiagonal is OEIS: A122045. The main diagonal is OEIS: A155585. The central column is OEIS: A099023. Row sums: 1, 1, −2, −5, 16, 61.... See OEIS: A163747. See the array beginning with 1, 1, 0, −2, 0, 16, 0 below.

The Akiyama–Tanigawa algorithm applied to OEIS: A046978 (n + 1) / OEIS: A016116(n) yields:

1. The first column is OEIS: A122045. Its binomial transform leads to:

The first row of this array is OEIS: A155585. The absolute values of the increasing antidiagonals are OEIS: A008280. The sum of the antidiagonals is −OEIS: A163747 (n + 1).

2. The second column is 1 1 −1 −5 5 61 −61 −1385 1385.... Its binomial transform yields:

The first row of this array is 1 2 2 −4 −16 32 272 544 −7936 15872 353792 −707584.... The absolute values of the second bisection are the double of the absolute values of the first bisection.

Consider the Akiyama-Tanigawa algorithm applied to OEIS: A046978 (n) / (OEIS: A158780 (n + 1) = abs(OEIS: A117575 (n)) + 1 = 1, 2, 2, ⁠3/2⁠, 1, ⁠3/4⁠, ⁠3/4⁠, ⁠7/8⁠, 1, ⁠17/16⁠, ⁠17/16⁠, ⁠33/32⁠....

The first column whose the absolute values are OEIS: A000111 could be the numerator of a trigonometric function.

OEIS: A163747 is an autosequence of the first kind (the main diagonal is OEIS: A000004). The corresponding array is:

The first two upper diagonals are −1 3 −24 402... = (−1)^{n + 1} × OEIS: A002832. The sum of the antidiagonals is 0 −2 0 10... = 2 × OEIS: A122045(n + 1).

−OEIS: A163982 is an autosequence of the second kind, like for instance OEIS: A164555 / OEIS: A027642. Hence the array:

The main diagonal, here 2 −2 8 −92..., is the double of the first upper one, here OEIS: A099023. The sum of the antidiagonals is 2 0 −4 0... = 2 × OEIS: A155585(n + 1). OEIS: A163747 − OEIS: A163982 = 2 × OEIS: A122045.

A combinatorial view: alternating permutations

Around 1880, three years after the publication of Seidel's algorithm, Désiré André proved a now classic result of combinatorial analysis. Looking at the first terms of the Taylor expansion of the trigonometric functions tan x and sec x, André made a discovery.

\begin{matrix} \tan x & = x + \frac{2 x^{3}}{3!} + \frac{16 x^{5}}{5!} + \frac{272 x^{7}}{7!} + \frac{7936 x^{9}}{9!} + \dots \\ \sec x & = 1 + \frac{x^{2}}{2!} + \frac{5 x^{4}}{4!} + \frac{61 x^{6}}{6!} + \frac{1385 x^{8}}{8!} + \frac{50521 x^{10}}{10!} + \dots \end{matrix}

The coefficients are the Euler zigzag numbers of odd and even index, respectively. In consequence the ordinary expansion of tan x + sec x has as coefficients the rational numbers S_n.

\tan x + \sec x = 1 + x + \frac{1}{2} x^{2} + \frac{1}{3} x^{3} + \frac{5}{24} x^{4} + \frac{2}{15} x^{5} + \frac{61}{720} x^{6} + \dots

André then succeeded by means of a recurrence argument to show that the alternating permutations of odd size are enumerated by the Euler numbers of odd index (also called tangent numbers) and the alternating permutations of even size by the Euler numbers of even index (also called secant numbers).

The arithmetic mean of the first and the second Bernoulli numbers are the associate Bernoulli numbers: B₀ = 1, B₁ = 0, B₂ = ⁠1/6⁠, B₃ = 0, B₄ = −⁠1/30⁠, OEIS: A176327 / OEIS: A027642. Via the second row of its inverse Akiyama–Tanigawa transform OEIS: A177427, they lead to Balmer series OEIS: A061037 / OEIS: A061038.

The Akiyama–Tanigawa algorithm applied to OEIS: A060819 (n + 4) / OEIS: A145979 (n) leads to the Bernoulli numbers OEIS: A027641 / OEIS: A027642, OEIS: A164555 / OEIS: A027642, or OEIS: A176327 OEIS: A176289 without B₁, named intrinsic Bernoulli numbers B_i(n).

Hence another link between the intrinsic Bernoulli numbers and the Balmer series via OEIS: A145979 (n).

OEIS: A145979 (n − 2) = 0, 2, 1, 6,... is a permutation of the non-negative numbers.

The terms of the first row are f(n) = ⁠1/2⁠ + ⁠1/n + 2⁠. 2, f(n) is an autosequence of the second kind. 3/2, f(n) leads by its inverse binomial transform to 3/2 −1/2 1/3 −1/4 1/5 ... = 1/2 + log 2.

Consider g(n) = 1/2 – 1 / (n+2) = 0, 1/6, 1/4, 3/10, 1/3. The Akiyama-Tanagiwa transforms gives:

0, g(n), is an autosequence of the second kind.

Euler OEIS: A198631 (n) / OEIS: A006519 (n + 1) without the second term (⁠1/2⁠) are the fractional intrinsic Euler numbers E_i(n) = 1, 0, −⁠1/4⁠, 0, ⁠1/2⁠, 0, −⁠17/8⁠, 0, ... The corresponding Akiyama transform is:

The first line is Eu(n). Eu(n) preceded by a zero is an autosequence of the first kind. It is linked to the Oresme numbers. The numerators of the second line are OEIS: A069834 preceded by 0. The difference table is:

Arithmetical properties of the Bernoulli numbers

The Bernoulli numbers can be expressed in terms of the Riemann zeta function as B_n = −nζ(1 − n) for integers n ≥ 0 provided for n = 0 the expression −nζ(1 − n) is understood as the limiting value and the convention B₁ = ⁠1/2⁠ is used. This intimately relates them to the values of the zeta function at negative integers. As such, they could be expected to have and do have deep arithmetical properties. For example, the Agoh–Giuga conjecture postulates that p is a prime number if and only if pB_{p − 1} is congruent to −1 modulo p. Divisibility properties of the Bernoulli numbers are related to the ideal class groups of cyclotomic fields by a theorem of Kummer and its strengthening in the Herbrand-Ribet theorem, and to class numbers of real quadratic fields by Ankeny–Artin–Chowla.

The Kummer theorems

The Bernoulli numbers are related to Fermat's Last Theorem (FLT) by Kummer's theorem, which says:

If the odd prime p does not divide any of the numerators of the Bernoulli numbers B₂, B₄, ..., B_{p − 3} then x^p + y^p + z^p = 0 has no solutions in nonzero integers.

Prime numbers with this property are called regular primes. Another classical result of Kummer are the following congruences.

Let p be an odd prime and b an even number such that p − 1 does not divide b. Then for any non-negative integer k

\frac{B_{k (p - 1) + b}}{k (p - 1) + b} \equiv \frac{B_{b}}{b} (\mod p) .

A generalization of these congruences goes by the name of p-adic continuity.

p-adic continuity

If b, m and n are positive integers such that m and n are not divisible by p − 1 and m ≡ n (mod p^{b − 1} (p − 1)), then

(1 - p^{m - 1}) \frac{B_{m}}{m} \equiv (1 - p^{n - 1}) \frac{B_{n}}{n} (\mod p^{b}) .

Since B_n = −nζ(1 − n), this can also be written

(1 - p^{- u}) ζ (u) \equiv (1 - p^{- v}) ζ (v) (\mod p^{b}),

where u = 1 − m and v = 1 − n, so that u and v are nonpositive and not congruent to 1 modulo p − 1. This tells us that the Riemann zeta function, with 1 − p^−s taken out of the Euler product formula, is continuous in the p-adic numbers on odd negative integers congruent modulo p − 1 to a particular a ≢ 1 mod (p − 1), and so can be extended to a continuous function ζ_p(s) for all p-adic integers $Z_{p},$ the p-adic zeta function.

Ramanujan's congruences

The following relations, due to Ramanujan, provide a method for calculating Bernoulli numbers that is more efficient than the one given by their original recursive definition:

(\binom{m + 3}{m}) B_{m} = {\begin{cases} \frac{m + 3}{3} - \sum_{j = 1}^{\frac{m}{6}} (\binom{m + 3}{m - 6 j}) B_{m - 6 j}, & if m \equiv 0 (\mod 6); \\ \frac{m + 3}{3} - \sum_{j = 1}^{\frac{m - 2}{6}} (\binom{m + 3}{m - 6 j}) B_{m - 6 j}, & if m \equiv 2 (\mod 6); \\ - \frac{m + 3}{6} - \sum_{j = 1}^{\frac{m - 4}{6}} (\binom{m + 3}{m - 6 j}) B_{m - 6 j}, & if m \equiv 4 (\mod 6) . \end{cases}

Von Staudt–Clausen theorem

The von Staudt–Clausen theorem was given by Karl Georg Christian von Staudt and Thomas Clausen independently in 1840. The theorem states that for every n > 0,

B_{2 n} + \sum_{(p - 1) ∣ 2 n} \frac{1}{p}

is an integer. The sum extends over all primes p for which p − 1 divides 2n.

A consequence of this is that the denominator of B_2n is given by the product of all primes p for which p − 1 divides 2n. In particular, these denominators are square-free and divisible by 6.

Why do the odd Bernoulli numbers vanish?

The sum

φ_{k} (n) = \sum_{i = 0}^{n} i^{k} - \frac{n^{k}}{2}

can be evaluated for negative values of the index n. Doing so will show that it is an odd function for even values of k, which implies that the sum has only terms of odd index. This and the formula for the Bernoulli sum imply that B_{2k + 1 − m} is 0 for m even and 2k + 1 − m > 1; and that the term for B₁ is cancelled by the subtraction. The von Staudt–Clausen theorem combined with Worpitzky's representation also gives a combinatorial answer to this question (valid for n > 1).

From the von Staudt–Clausen theorem it is known that for odd n > 1 the number 2B_n is an integer. This seems trivial if one knows beforehand that the integer in question is zero. However, by applying Worpitzky's representation one gets

2 B_{n} = \sum_{m = 0}^{n} (- 1)^{m} \frac{2}{m + 1} m! {\binom{n + 1}{m + 1}} = 0 (n > 1 is odd)

as a sum of integers, which is not trivial. Here a combinatorial fact comes to surface which explains the vanishing of the Bernoulli numbers at odd index. Let S_n,m be the number of surjective maps from {1, 2, ..., n} to {1, 2, ..., m}, then S_n,m = m!{ⁿ
_m}. The last equation can only hold if

\sum_{odd m = 1}^{n - 1} \frac{2}{m^{2}} S_{n, m} = \sum_{even m = 2}^{n} \frac{2}{m^{2}} S_{n, m} (n > 2 is even) .

This equation can be proved by induction. The first two examples of this equation are

n = 4: 2 + 8 = 7 + 3,

n = 6: 2 + 120 + 144 = 31 + 195 + 40.

Thus the Bernoulli numbers vanish at odd index because some non-obvious combinatorial identities are embodied in the Bernoulli numbers.

A restatement of the Riemann hypothesis

The connection between the Bernoulli numbers and the Riemann zeta function is strong enough to provide an alternate formulation of the Riemann hypothesis (RH) which uses only the Bernoulli numbers. In fact Marcel Riesz proved that the RH is equivalent to the following assertion:

For every ε > ⁠1/4⁠ there exists a constant C_ε > 0 (depending on ε) such that |R(x)| < C_εx^ε as x → ∞.

Here R(x) is the Riesz function

R (x) = 2 \sum_{k = 1}^{\infty} \frac{k^{\bar{k}} x^{k}}{(2 π)^{2 k} (\frac{B_{2 k}}{2 k})} = 2 \sum_{k = 1}^{\infty} \frac{k^{\bar{k}} x^{k}}{(2 π)^{2 k} β_{2 k}} .

n^k denotes the rising factorial power in the notation of D. E. Knuth. The numbers β_n = ⁠B_n/n⁠ occur frequently in the study of the zeta function and are significant because β_n is a p-integer for primes p where p − 1 does not divide n. The β_n are called divided Bernoulli numbers.

Generalized Bernoulli numbers

The generalized Bernoulli numbers are certain algebraic numbers, defined similarly to the Bernoulli numbers, that are related to special values of Dirichlet L-functions in the same way that Bernoulli numbers are related to special values of the Riemann zeta function.

Let χ be a Dirichlet character modulo f. The generalized Bernoulli numbers attached to χ are defined by

\sum_{a = 1}^{f} χ (a) \frac{t e^{a t}}{e^{f t} - 1} = \sum_{k = 0}^{\infty} B_{k, χ} \frac{t^{k}}{k!} .

Apart from the exceptional B_1,1 = ⁠1/2⁠, we have, for any Dirichlet character χ, that B_k,χ = 0 if χ(−1) ≠ (−1)^k.

Generalizing the relation between Bernoulli numbers and values of the Riemann zeta function at non-positive integers, one has the for all integers k ≥ 1:

L (1 - k, χ) = - \frac{B_{k, χ}}{k},

where L(s,χ) is the Dirichlet L-function of χ.

Eisenstein–Kronecker number

Eisenstein–Kronecker numbers are an analogue of the generalized Bernoulli numbers for imaginary quadratic fields. They are related to critical L-values of Hecke characters.

Appendix

Assorted identities

Umbral calculus gives a compact form of Bernoulli's formula, by using an abstract symbol B,
$S_{m} (n) = \frac{1}{m + 1} ((B + n)^{m + 1} - B_{m + 1}),$

where the symbol B^k that appears during binomial expansion of the parenthesized term is to be replaced by the Bernoulli number B_k (and B₁ = +⁠1/2⁠). More suggestively and mnemonically, this may be written as

$S_{m} (n) = \int_{0}^{n} (B + x)^{m} d x .$

Other Bernoulli identities can be written compactly with this symbol. For example,

$(1 - 2 B)^{m} = (2 - 2^{m}) B_{m} .$
Let n be non-negative and even. Then
$ζ (n) = \frac{(- 1)^{\frac{n}{2} - 1} B_{n} (2 π)^{n}}{2 (n!)} .$
The nth cumulant of the uniform probability distribution on the interval [−1, 0] is ⁠B_n/n⁠.
For n ≥ 1, B_n is given by the determinant of an (n + 1) × (n + 1) matrix. Namely,
$B_{n} = n! | \begin{matrix} 1 & 0 & \dots & 0 & 1 \\ \frac{1}{2!} & 1 & \dots & 0 & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ & ⋮ \\ \frac{1}{n!} & \frac{1}{(n - 1)!} & \dots & 1 & 0 \\ \frac{1}{(n + 1)!} & \frac{1}{n!} & \dots & \frac{1}{2!} & 0 \end{matrix} | .$
So, the determinant is σ_n(1), the Stirling polynomial at x = 1.
The even-numbered Bernoulli number B_2n is given by the determinant of an (n + 1) × (n + 1) matrix. Namely,
$B_{2 n} = - \frac{(2 n)!}{2^{2 n} - 2} | \begin{matrix} 1 & 0 & 0 & \dots & 0 & 1 \\ \frac{1}{3!} & 1 & 0 & \dots & 0 & 0 \\ \frac{1}{5!} & \frac{1}{3!} & 1 & \dots & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋱ & ⋮ & ⋮ \\ \frac{1}{(2 n + 1)!} & \frac{1}{(2 n - 1)!} & \frac{1}{(2 n - 3)!} & \dots & \frac{1}{3!} & 0 \end{matrix} | .$
Let n ≥ 1. Then
$\frac{1}{n} \sum_{k = 1}^{n} (\binom{n}{k}) B_{k} B_{n - k} + B_{n - 1} = - B_{n} .$
Let n ≥ 1. Then
$\sum_{k = 0}^{n} (\binom{n + 1}{k}) (n + k + 1) B_{n + k} = 0.$
Let n ≥ 0. Then (Leopold Kronecker 1883)
$B_{n} = - \sum_{k = 1}^{n + 1} \frac{(- 1)^{k}}{k} (\binom{n + 1}{k}) \sum_{j = 1}^{k} j^{n} .$
Let n ≥ 1 and m ≥ 1. Then
$(- 1)^{m} \sum_{r = 0}^{m} (\binom{m}{r}) B_{n + r} = (- 1)^{n} \sum_{s = 0}^{n} (\binom{n}{s}) B_{m + s} .$
Let n ≥ 4 and H_n denote the nth harmonic number, that is,
$H_{n} = \sum_{k = 1}^{n} k^{- 1} .$

Then (H. Miki 1978)

$\frac{n}{2} \sum_{k = 2}^{n - 2} \frac{B_{n - k}}{n - k} \frac{B_{k}}{k} - \sum_{k = 2}^{n - 2} (\binom{n}{k}) \frac{B_{n - k}}{n - k} B_{k} = H_{n} B_{n} .$
Let n ≥ 4. Then (Yuri Matiyasevich 1997)
$(n + 2) \sum_{k = 2}^{n - 2} B_{k} B_{n - k} - 2 \sum_{l = 2}^{n - 2} (\binom{n + 2}{l}) B_{l} B_{n - l} = n (n + 1) B_{n} .$
The Faber–Pandharipande–Zagier–Gessel identity states that for n ≥ 1,
$\frac{n}{2} (B_{n - 1} (x) + \sum_{k = 1}^{n - 1} \frac{B_{k} (x)}{k} \frac{B_{n - k} (x)}{n - k}) - \sum_{k = 0}^{n - 1} (\binom{n}{k}) \frac{B_{n - k}}{n - k} B_{k} (x) = H_{n - 1} B_{n} (x) .$
Choosing x = 0 or x = 1 results in the Bernoulli number identity.
For n ≥ 0 if B₁ = ⁠1/2⁠, or n ≥ 1 if B₁ = −⁠1/2⁠,
$\sum_{k = 0}^{n} (\binom{n}{k}) \frac{B_{k}}{n - k + 2} = \frac{B_{n + 1}}{n + 1} .$
Let n ≥ 0. Then
$- 1 + \sum_{k = 0}^{n} (\binom{n}{k}) \frac{2^{n - k + 1}}{n - k + 1} B_{k} (1) = 2^{n}$

and

$- 1 + \sum_{k = 0}^{n} (\binom{n}{k}) \frac{2^{n - k + 1}}{n - k + 1} B_{k} (0) = δ_{n, 0} .$
A reciprocity relation of M. B. Gelfand states that
$(- 1)^{m + 1} \sum_{j = 0}^{k} (\binom{k}{j}) \frac{B_{m + 1 + j}}{m + 1 + j} + (- 1)^{k + 1} \sum_{j = 0}^{m} (\binom{m}{j}) \frac{B_{k + 1 + j}}{k + 1 + j} = \frac{k! m!}{(k + m + 1)!} .$