Narcissistic number

Definition

Let ${\displaystyle n}$ be a natural number. We define the narcissistic function for base ${\displaystyle b>1}$ ${\displaystyle F_b:\mathbb{N}\to \mathbb{N}}$ to be the following:

${\displaystyle F_b(n)=\sum _{i=0}^{k-1}{d}_i^k.}$

where ${\displaystyle k=\lfloor {\log }_{b}n\rfloor +1}$ is the number of digits in the number in base ${\displaystyle b}$, and

${\displaystyle d_i=\frac{nmod{b}^{i+1}-n{modb}^i}{b^i}}$

is the value of each digit of the number. A natural number ${\displaystyle n}$ is a narcissistic number if it is a fixed point for ${\displaystyle F_b}$, which occurs if ${\displaystyle F_b(n)=n}$. The natural numbers ${\displaystyle 0\le n<b}$ are trivial narcissistic numbers for all ${\displaystyle b}$, all other narcissistic numbers are nontrivial narcissistic numbers.

For example, the number 153 in base ${\displaystyle b=10}$ is a narcissistic number, because ${\displaystyle k=3}$ and ${\displaystyle 153=1^3+5^3+3^3}$.

A natural number ${\displaystyle n}$ is a sociable narcissistic number if it is a periodic point for ${\displaystyle F_b}$, where ${\displaystyle F_b^p(n)=n}$ for a positive integer ${\displaystyle p}$ (here ${\displaystyle F_b^p}$ is the ${\displaystyle p}$th iterate of ${\displaystyle F_b}$), and forms a cycle of period ${\displaystyle p}$. A narcissistic number is a sociable narcissistic number with ${\displaystyle p=1}$, and an amicable narcissistic number is a sociable narcissistic number with ${\displaystyle p=2}$.

All natural numbers ${\displaystyle n}$ are preperiodic points for ${\displaystyle F_b}$, regardless of the base. This is because for any given digit count ${\displaystyle k}$, the minimum possible value of ${\displaystyle n}$ is ${\displaystyle b^{k-1}}$, the maximum possible value of ${\displaystyle n}$ is ${\displaystyle b^k-1\le b^k}$, and the narcissistic function value is ${\displaystyle F_b(n)=k(b-1{)}^k}$. Thus, any narcissistic number must satisfy the inequality ${\displaystyle b^{k-1}\le k(b-1{)}^k\le b^k}$. Multiplying all sides by ${\displaystyle \frac{b}{(b-1{)}^k}}$, we get ${\displaystyle {\left(\frac{b}{b-1}\right)}^k\le bk\le b{\left(\frac{b}{b-1}\right)}^k}$, or equivalently, ${\displaystyle k\le {\left(\frac{b}{b-1}\right)}^k\le bk}$. Since ${\displaystyle \frac{b}{b-1}\ge 1}$, this means that there will be a maximum value ${\displaystyle k}$ where ${\displaystyle {\left(\frac{b}{b-1}\right)}^k\le bk}$, because of the exponential nature of ${\displaystyle {\left(\frac{b}{b-1}\right)}^k}$ and the linearity of ${\displaystyle bk}$. Beyond this value ${\displaystyle k}$, ${\displaystyle F_b(n)\le n}$ always. Thus, there are a finite number of narcissistic numbers, and any natural number is guaranteed to reach a periodic point or a fixed point less than ${\displaystyle b^k-1}$, making it a preperiodic point. Setting ${\displaystyle b}$ equal to 10 shows that the largest narcissistic number in base 10 must be less than ${\displaystyle {10}^{60}}$.

The number of iterations ${\displaystyle i}$ needed for ${\displaystyle F_b^i(n)}$ to reach a fixed point is the narcissistic function's persistence of ${\displaystyle n}$, and undefined if it never reaches a fixed point.

A base ${\displaystyle b}$ has at least one two-digit narcissistic number if and only if ${\displaystyle b^2+1}$ is not prime, and the number of two-digit narcissistic numbers in base ${\displaystyle b}$ equals ${\displaystyle \tau (b^2+1)-2}$, where ${\displaystyle \tau (n)}$ is the number of positive divisors of ${\displaystyle n}$.

Every base ${\displaystyle b\ge 3}$ that is not a multiple of nine has at least one three-digit narcissistic number. The bases that do not are

2, 72, 90, 108, 153, 270, 423, 450, 531, 558, 630, 648, 738, 1044, 1098, 1125, 1224, 1242, 1287, 1440, 1503, 1566, 1611, 1620, 1800, 1935, ... (sequence A248970 in the OEIS)

There are only 88 narcissistic numbers in base 10, of which the largest is

115,132,219,018,763,992,565,095,597,973,971,522,401

with 39 digits.