Kaprekar number · Lablypedia

Definition and properties

Let $n$ be a natural number. Then the Kaprekar function for base $b > 1$ and power $p > 0$ $F_{p, b} : N \to N$ is defined to be the following:

F_{p, b} (n) = α + β

where $β = n^{2} {mod b}^{p}$ and

α = \frac{n^{2} - β}{b^{p}}

A natural number $n$ is a $p$ -Kaprekar number if it is a fixed point for $F_{p, b}$ , which occurs if $F_{p, b} (n) = n$ . $0$ and $1$ are trivial Kaprekar numbers for all $b$ and $p$ , all other Kaprekar numbers are nontrivial Kaprekar numbers.

The earlier example of 45 satisfies this definition with $b = 10$ and $p = 2$ , because

β = n^{2} {mod b}^{p} = 45^{2} mod 1 0^{2} = 25

α = \frac{n^{2} - β}{b^{p}} = \frac{45^{2} - 25}{10^{2}} = 20

F_{2, 10} (45) = α + β = 20 + 25 = 45

A natural number $n$ is a sociable Kaprekar number if it is a periodic point for $F_{p, b}$ , where $F_{p, b}^{k} (n) = n$ for a positive integer $k$ (where $F_{p, b}^{k}$ is the $k$ th iterate of $F_{p, b}$ ), and forms a cycle of period $k$ . A Kaprekar number is a sociable Kaprekar number with $k = 1$ , and a amicable Kaprekar number is a sociable Kaprekar number with $k = 2$ .

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

There are only a finite number of $p$ -Kaprekar numbers and cycles for a given base $b$ , because if $n = b^{p} + m$ , where $m > 0$ then

\begin{matrix} n^{2} & = (b^{p} + m)^{2} \\ = b^{2 p} + 2 m b^{p} + m^{2} \\ = (b^{p} + 2 m) b^{p} + m^{2} \end{matrix}

and $β = m^{2}$ , $α = b^{p} + 2 m$ , and $F_{p, b} (n) = b^{p} + 2 m + m^{2} = n + (m^{2} + m) > n$ . Only when $n \leq b^{p}$ do Kaprekar numbers and cycles exist.

If $d$ is any divisor of $p$ , then $n$ is also a $p$ -Kaprekar number for base $b^{p}$ .

In base $b = 2$ , all even perfect numbers are Kaprekar numbers. More generally, any numbers of the form $2^{n} (2^{n + 1} - 1)$ or $2^{n} (2^{n + 1} + 1)$ for natural number $n$ are Kaprekar numbers in base 2.

Set-theoretic definition and unitary divisors

The set $K (N)$ for a given integer $N$ can be defined as the set of integers $X$ for which there exist natural numbers $A$ and $B$ satisfying the Diophantine equation

X^{2} = A N + B

, where

0 \leq B < N

X = A + B

An $n$ -Kaprekar number for base $b$ is then one which lies in the set $K (b^{n})$ .

It was shown in 2000 that there is a bijection between the unitary divisors of $N - 1$ and the set $K (N)$ defined above. Let $Inv (a, c)$ denote the multiplicative inverse of $a$ modulo $c$ , namely the least positive integer $m$ such that $a m = 1 mod c$ , and for each unitary divisor $d$ of $N - 1$ let $e = \frac{N - 1}{d}$ and $ζ (d) = d Inv (d, e)$ . Then the function $ζ$ is a bijection from the set of unitary divisors of $N - 1$ onto the set $K (N)$ . In particular, a number $X$ is in the set $K (N)$ if and only if $X = d Inv (d, e)$ for some unitary divisor $d$ of $N - 1$ .

The numbers in $K (N)$ occur in complementary pairs, $X$ and $N - X$ . If $d$ is a unitary divisor of $N - 1$ then so is $e = \frac{N - 1}{d}$ , and if $X = d Inv (d, e)$ then $N - X = e Inv (e, d)$ .

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Kaprekar numbers for F p , b {\displaystyle F_{p,b}}

b = 4k + 3 and p = 2n + 1

Let $k$ and $n$ be natural numbers, the number base $b = 4 k + 3 = 2 (2 k + 1) + 1$ , and $p = 2 n + 1$ . Then:

$X_{1} = \frac{b^{p} - 1}{2} = (2 k + 1) \sum_{i = 0}^{p - 1} b^{i}$ is a Kaprekar number.

Proof

Let

$\begin{matrix} X_{1} & = \frac{b^{p} - 1}{2} \\ = \frac{b - 1}{2} \sum_{i = 0}^{p - 1} b^{i} \\ = \frac{4 k + 3 - 1}{2} \sum_{i = 0}^{2 n + 1 - 1} b^{i} \\ = (2 k + 1) \sum_{i = 0}^{2 n} b^{i} \end{matrix}$

Then,

$\begin{matrix} X_{1}^{2} & = {(\frac{b^{p} - 1}{2})}^{2} \\ = \frac{b^{2 p} - 2 b^{p} + 1}{4} \\ = \frac{b^{p} (b^{p} - 2) + 1}{4} \\ = \frac{(4 k + 3)^{2 n + 1} (b^{p} - 2) + 1}{4} \\ = \frac{(4 k + 3)^{2 n} (b^{p} - 2) (4 k + 4) - (4 k + 3)^{2 n} (b^{p} - 2) + 1}{4} \\ = \frac{- (4 k + 3)^{2 n} (b^{p} - 2) + 1}{4} + (k + 1) (4 k + 3)^{2 n} (b^{p} - 2) \\ = \frac{- (4 k + 3)^{2 n - 1} (b^{p} - 2) (4 k + 4) + (4 k + 3)^{2 n - 1} (b^{p} - 2) + 1}{4} + (k + 1) b^{2 n} (b^{2 n + 1} - 2) \\ = \frac{(4 k + 3)^{2 n - 1} (b^{p} - 2) + 1}{4} + (k + 1) b^{2 n} (b^{p} - 2) - (k + 1) b^{2 n - 1} (b^{2 n + 1} - 2) \\ = \frac{(4 k + 3)^{p - 2} (b^{p} - 2) + 1}{4} + \sum_{i = p - 2}^{p - 1} (- 1)^{i} (k + 1) b^{i} (b^{p} - 2) \\ = \frac{(4 k + 3)^{p - 2} (b^{p} - 2) + 1}{4} + (b^{p} - 2) (k + 1) \sum_{i = p - 2}^{p - 1} (- 1)^{i} b^{i} \\ = \frac{(4 k + 3)^{1} (b^{p} - 2) + 1}{4} + (b^{p} - 2) (k + 1) \sum_{i = 1}^{p - 1} (- 1)^{i} b^{i} \\ = \frac{- (b^{p} - 2) + 1}{4} + (b^{p} - 2) (k + 1) \sum_{i = 0}^{p - 1} (- 1)^{i} b^{i} \\ = (b^{p} - 2) (k + 1) (\sum_{i = 0}^{2 n} (- 1)^{i} b^{i}) + \frac{- b^{2 n + 1} + 3}{4} \\ = (b^{p} - 2) (k + 1) (\sum_{i = 0}^{2 n} (- 1)^{i} b^{i}) + \frac{- 4 b^{2 n + 1} + 3 b^{2 n + 1} + 3}{4} \\ = (b^{p} - 2) (k + 1) (\sum_{i = 0}^{2 n} (- 1)^{i} b^{i}) - b^{p} + \frac{3 b^{2 n + 1} + 3}{4} \\ = (b^{p} - 2) (k + 1) (\sum_{i = 0}^{2 n} (- 1)^{i} b^{i}) - b^{p} + \frac{3 (4 k + 3)^{p - 2} + 3}{4} + 3 (k + 1) \sum_{i = p - 2}^{p - 1} (- 1)^{i} b^{i} \\ = (b^{p} - 2) (k + 1) (\sum_{i = 0}^{2 n} (- 1)^{i} b^{i}) - b^{p} + \frac{3 (4 k + 3)^{1} + 3}{4} + 3 (k + 1) \sum_{i = 1}^{p - 1} (- 1)^{i} b^{i} \\ = (b^{p} - 2) (k + 1) (\sum_{i = 0}^{2 n} (- 1)^{i} b^{i}) - b^{p} + \frac{- 3 + 3}{4} + 3 (k + 1) \sum_{i = 0}^{p - 1} (- 1)^{i} b^{i} \\ = (b^{p} - 2) (k + 1) (\sum_{i = 0}^{2 n} (- 1)^{i} b^{i}) + 3 (k + 1) (\sum_{i = 0}^{2 n} (- 1)^{i} b^{i}) - b^{p} \\ = (b^{p} - 2 + 3) (k + 1) (\sum_{i = 0}^{2 n} (- 1)^{i} b^{i}) - b^{p} \\ = (b^{p} + 1) (k + 1) (\sum_{i = 0}^{2 n} (- 1)^{i} b^{i}) - b^{p} \\ = (b^{p} + 1) (- 1 + (k + 1) \sum_{i = 0}^{2 n} (- 1)^{i} b^{i}) + 1 \\ = (b^{p} + 1) (k + (k + 1) \sum_{i = 1}^{2 n} (- 1)^{i} b^{i}) + 1 \\ = (b^{p} + 1) (k + (k + 1) \sum_{i = 1}^{n} b^{2 i} - b^{2 i - 1}) + 1 \\ = (b^{p} + 1) (k + (k + 1) \sum_{i = 1}^{n} (b - 1) b^{2 i - 1}) + 1 \\ = (b^{p} + 1) (k + \sum_{i = 1}^{n} ((k + 1) b - k - 1) b^{2 i - 1}) + 1 \\ = (b^{p} + 1) (k + \sum_{i = 1}^{n} (k b + (4 k + 3) - k - 1) b^{2 i - 1}) + 1 \\ = (b^{p} + 1) (k + \sum_{i = 1}^{n} (k b + (3 k + 2)) b^{2 i - 1}) + 1 \\ = b^{p} (k + \sum_{i = 1}^{n} (k b + (3 k + 2)) b^{2 i - 1}) + (k + 1 + \sum_{i = 1}^{n} (k b + (3 k + 2)) b^{2 i - 1}) \end{matrix}$

The two numbers $α$ and $β$ are

β = X_{1}^{2} {mod b}^{p} = k + 1 + \sum_{i = 1}^{n} (k b + (3 k + 2)) b^{2 i - 1}

α = \frac{X_{1}^{2} - β}{b^{p}} = k + \sum_{i = 1}^{n} (k b + (3 k + 2)) b^{2 i - 1}

and their sum is

$\begin{matrix} α + β & = (k + \sum_{i = 1}^{n} (k b + (3 k + 2)) b^{2 i - 1}) + (k + 1 + \sum_{i = 1}^{n} (k b + (3 k + 2)) b^{2 i - 1}) \\ = 2 k + 1 + \sum_{i = 1}^{n} ((2 k) b + 2 (3 k + 2)) b^{2 i - 1} \\ = 2 k + 1 + \sum_{i = 1}^{n} ((2 k) b + (6 k + 4)) b^{2 i - 1} \\ = 2 k + 1 + \sum_{i = 1}^{n} ((2 k) b + (4 k + 3)) b^{2 i - 1} + (2 k + 1) b^{2 i - 1} \\ = 2 k + 1 + \sum_{i = 1}^{n} ((2 k + 1) b) b^{2 i - 1} + (2 k + 1) b^{2 i - 1} \\ = 2 k + 1 + \sum_{i = 1}^{n} (2 k + 1) b^{2 i} + (2 k + 1) b^{2 i - 1} \\ = 2 k + 1 + \sum_{i = 1}^{2 n} (2 k + 1) b^{i} \\ = \sum_{i = 0}^{2 n} (2 k + 1) b^{i} \\ = (2 k + 1) \sum_{i = 0}^{2 n} b^{i} & = X_{1} \end{matrix}$

Thus, $X_{1}$ is a Kaprekar number.

$X_{2} = \frac{b^{p} + 1}{2} = X_{1} + 1$ is a Kaprekar number for all natural numbers $n$ .

Proof

Let

$\begin{matrix} X_{2} & = \frac{b^{2 n + 1} + 1}{2} \\ = \frac{b^{2 n + 1} - 1}{2} + 1 \\ = X_{1} + 1 \end{matrix}$

Then,

$\begin{matrix} X_{2}^{2} & = (X_{1} + 1)^{2} \\ = X_{1}^{2} + 2 X_{1} + 1 \\ = X_{1}^{2} + 2 X_{1} + 1 \\ = b^{p} (k + \sum_{i = 1}^{n} (k b + (3 k + 2)) b^{2 i - 1}) + (k + 1 + \sum_{i = 1}^{n} (k b + (3 k + 2)) b^{2 i - 1}) + b^{p} - 1 + 1 \\ = b^{p} (k + 1 + \sum_{i = 1}^{n} (k b + (3 k + 2)) b^{2 i - 1}) + (k + 1 + \sum_{i = 1}^{n} (k b + (3 k + 2)) b^{2 i - 1}) \end{matrix}$

The two numbers $α$ and $β$ are

β = X_{2}^{2} {mod b}^{p} = k + 1 + \sum_{i = 1}^{n} (k b + (3 k + 2)) b^{2 i - 1}

α = \frac{X_{2}^{2} - β}{b^{p}} = k + 1 + \sum_{i = 1}^{n} (k b + (3 k + 2)) b^{2 i - 1}

and their sum is

$\begin{matrix} α + β & = (k + 1 + \sum_{i = 1}^{n} (k b + (3 k + 2)) b^{2 i - 1}) + (k + 1 + \sum_{i = 1}^{n} (k b + (3 k + 2)) b^{2 i - 1}) \\ = 2 k + 2 + \sum_{i = 1}^{n} ((2 k) b + 2 (3 k + 2)) b^{2 i - 1} \\ = 2 k + 2 + \sum_{i = 1}^{n} ((2 k) b + (6 k + 4)) b^{2 i - 1} \\ = 2 k + 2 + \sum_{i = 1}^{n} ((2 k) b + (4 k + 3)) b^{2 i - 1} + (2 k + 1) b^{2 i - 1} \\ = 2 k + 2 + \sum_{i = 1}^{n} ((2 k + 1) b) b^{2 i - 1} + (2 k + 1) b^{2 i - 1} \\ = 2 k + 2 + \sum_{i = 1}^{n} (2 k + 1) b^{2 i} + (2 k + 1) b^{2 i - 1} \\ = 2 k + 2 + \sum_{i = 1}^{2 n} (2 k + 1) b^{i} \\ = 1 + \sum_{i = 0}^{2 n} (2 k + 1) b^{i} \\ = 1 + (2 k + 1) \sum_{i = 0}^{2 n} b^{i} \\ = 1 + X_{1} \\ = X_{2} \end{matrix}$

Thus, $X_{2}$ is a Kaprekar number.

b = m2k + m + 1 and p = mn + 1

Let $m$ , $k$ , and $n$ be natural numbers, the number base $b = m^{2} k + m + 1$ , and the power $p = m n + 1$ . Then:

$X_{1} = \frac{b^{p} - 1}{m} = (m k + 1) \sum_{i = 0}^{p - 1} b^{i}$ is a Kaprekar number.
$X_{2} = \frac{b^{p} + m - 1}{m} = X_{1} + 1$ is a Kaprekar number.

b = m2k + m + 1 and p = mn + m − 1

Let $m$ , $k$ , and $n$ be natural numbers, the number base $b = m^{2} k + m + 1$ , and the power $p = m n + m - 1$ . Then:

$X_{1} = \frac{m (b^{p} - 1)}{4} = (m - 1) (m k + 1) \sum_{i = 0}^{p - 1} b^{i}$ is a Kaprekar number.
$X_{2} = \frac{m b^{p} + 1}{4} = X_{3} + 1$ is a Kaprekar number.

b = m2k + m2 − m + 1 and p = mn + 1

Let $m$ , $k$ , and $n$ be natural numbers, the number base $b = m^{2} k + m^{2} - m + 1$ , and the power $p = m n + m - 1$ . Then:

$X_{1} = \frac{(m - 1) (b^{p} - 1)}{m} = (m - 1) (m k + 1) \sum_{i = 0}^{p - 1} b^{i}$ is a Kaprekar number.
$X_{2} = \frac{(m - 1) b^{p} + 1}{m} = X_{1} + 1$ is a Kaprekar number.

b = m2k + m2 − m + 1 and p = mn + m − 1