عدد كاپريكار

يجب ألا يـُخلـَط بينها وبين ثابت كاپريكار.

In mathematics, a natural number in a given number base is a ${\displaystyle{\displaystyle p}}$-Kaprekar number if the representation of its square in that base can be split into two parts, where the second part has ${\displaystyle{\displaystyle p}}$ digits, that add up to the original number. الأعداد مسماة على اسم د. ر. كاپريكار.

 التعريف والخصائص

Let ${\displaystyle{\displaystyle n}}$ be a natural number. We define the Kaprekar function for base ${\displaystyle{\displaystyle b>1}}$ and power ${\displaystyle{\displaystyle p>0}}$ ${\displaystyle{\displaystyle F_{p,b}:\mathbb{N}\rightarrow\mathbb{N}}}$ to be the following:

${\displaystyle{\displaystyle F_{p,b}(n)=\alpha+\beta}}$,

where ${\displaystyle{\displaystyle\beta=n^{2}{\bmod{b}}^{p}}}$ and

${\displaystyle{\displaystyle\alpha={\frac{n^{2}-\beta}{b^{p}}}}}$

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

For example, in base 10, 45 is a 2-Kaprekar number, because

${\displaystyle{\displaystyle\beta=n^{2}{\bmod{b}}^{p}=45^{2}{\bmod{1}}0^{2}=25}}$

${\displaystyle{\displaystyle\alpha={\frac{n^{2}-\beta}{b^{p}}}={\frac{45^{2}-25}{10^{2}}}=20}}$

${\displaystyle{\displaystyle F_{2,10}(45)=\alpha+\beta=20+25=45}}$

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

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

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

${\displaystyle{\displaystyle{\begin{aligned} \displaystyle n^{2}&\displaystyle=(b^{p}+m)^{2}\\ &\displaystyle=b^{2p}+2mb^{p}+m^{2}\\ &\displaystyle=(b^{p}+2m)b^{p}+m^{2}\\ \end{aligned}}}}$

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

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

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

 التعريف حسب نظرية الفئات وقواسم الوحدة

We can define the set ${\displaystyle{\displaystyle K(N)}}$ for a given integer ${\displaystyle{\displaystyle N}}$ as the set of integers ${\displaystyle{\displaystyle X}}$ for which there exist natural numbers ${\displaystyle{\displaystyle A}}$ and ${\displaystyle{\displaystyle B}}$ satisfying the Diophantine equation

${\displaystyle{\displaystyle X^{2}=AN+B}}$, where ${\displaystyle{\displaystyle 0\leq B<N}}$

${\displaystyle{\displaystyle X=A+B}}$

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

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

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

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

 b = 4k + 3 and p = 2n + 1

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

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

Proof

Let

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

Then,

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

The two numbers ${\displaystyle{\displaystyle\alpha}}$ and ${\displaystyle{\displaystyle\beta}}$ are

${\displaystyle{\displaystyle\beta=X_{1}^{2}{\bmod{b}}^{p}=k+1+\sum_{i=1}^{n}(kb+(3k+2))b^{2i-1}}}$

${\displaystyle{\displaystyle\alpha={\frac{X_{1}^{2}-\beta}{b^{p}}}=k+\sum_{i=1}^{n}(kb+(3k+2))b^{2i-1}}}$

and their sum is

${\displaystyle{\displaystyle{\begin{aligned} \displaystyle\alpha+\beta&\displaystyle=\left(k+\sum_{i=1}^{n}(kb+(3k+2))b^{2i-1}\right)+\left(k+1+\sum_{i=1}^{n}(kb+(3k+2))b^{2i-1}\right)\\ &\displaystyle=2k+1+\sum_{i=1}^{n}((2k)b+2(3k+2))b^{2i-1}\\ &\displaystyle=2k+1+\sum_{i=1}^{n}((2k)b+(6k+4))b^{2i-1}\\ &\displaystyle=2k+1+\sum_{i=1}^{n}((2k)b+(4k+3))b^{2i-1}+(2k+1)b^{2i-1}\\ &\displaystyle=2k+1+\sum_{i=1}^{n}((2k+1)b)b^{2i-1}+(2k+1)b^{2i-1}\\ &\displaystyle=2k+1+\sum_{i=1}^{n}(2k+1)b^{2i}+(2k+1)b^{2i-1}\\ &\displaystyle=2k+1+\sum_{i=1}^{2n}(2k+1)b^{i}\\ &\displaystyle=\sum_{i=0}^{2n}(2k+1)b^{i}\\ &\displaystyle=(2k+1)\sum_{i=0}^{2n}b^{i}&\displaystyle=X_{1}\\ \end{aligned}}}}$

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

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

Proof

Let

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

Then,

${\displaystyle{\displaystyle{\begin{aligned} \displaystyle X_{2}^{2}&\displaystyle=(X_{1}+1)^{2}\\ &\displaystyle=X_{1}^{2}+2X_{1}+1\\ &\displaystyle=X_{1}^{2}+2X_{1}+1\\ &\displaystyle=b^{p}\left(k+\sum_{i=1}^{n}(kb+(3k+2))b^{2i-1}\right)+\left(k+1+\sum_{i=1}^{n}(kb+(3k+2))b^{2i-1}\right)+b^{p}-1+1\\ &\displaystyle=b^{p}\left(k+1+\sum_{i=1}^{n}(kb+(3k+2))b^{2i-1}\right)+\left(k+1+\sum_{i=1}^{n}(kb+(3k+2))b^{2i-1}\right)\end{aligned}}}}$

The two numbers ${\displaystyle{\displaystyle\alpha}}$ and ${\displaystyle{\displaystyle\beta}}$ are

${\displaystyle{\displaystyle\beta=X_{2}^{2}{\bmod{b}}^{p}=k+1+\sum_{i=1}^{n}(kb+(3k+2))b^{2i-1}}}$

${\displaystyle{\displaystyle\alpha={\frac{X_{2}^{2}-\beta}{b^{p}}}=k+1+\sum_{i=1}^{n}(kb+(3k+2))b^{2i-1}}}$

and their sum is

${\displaystyle{\displaystyle{\begin{aligned} \displaystyle\alpha+\beta&\displaystyle=\left(k+1+\sum_{i=1}^{n}(kb+(3k+2))b^{2i-1}\right)+\left(k+1+\sum_{i=1}^{n}(kb+(3k+2))b^{2i-1}\right)\\ &\displaystyle=2k+2+\sum_{i=1}^{n}((2k)b+2(3k+2))b^{2i-1}\\ &\displaystyle=2k+2+\sum_{i=1}^{n}((2k)b+(6k+4))b^{2i-1}\\ &\displaystyle=2k+2+\sum_{i=1}^{n}((2k)b+(4k+3))b^{2i-1}+(2k+1)b^{2i-1}\\ &\displaystyle=2k+2+\sum_{i=1}^{n}((2k+1)b)b^{2i-1}+(2k+1)b^{2i-1}\\ &\displaystyle=2k+2+\sum_{i=1}^{n}(2k+1)b^{2i}+(2k+1)b^{2i-1}\\ &\displaystyle=2k+2+\sum_{i=1}^{2n}(2k+1)b^{i}\\ &\displaystyle=1+\sum_{i=0}^{2n}(2k+1)b^{i}\\ &\displaystyle=1+(2k+1)\sum_{i=0}^{2n}b^{i}\\ &\displaystyle=1+X_{1}\\ &\displaystyle=X_{2}\end{aligned}}}}$

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

b = m²k + m + 1 and p = mn + 1

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

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

 b = m²k + m + 1 and p = mn + m − 1

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

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

 b = m²k + m² − m + 1 and p = mn + 1

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

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

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

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

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

Kaprekar numbers and cycles of ${\displaystyle{\displaystyle F_{p,b}}}$ for specific ${\displaystyle{\displaystyle p}}$, ${\displaystyle{\displaystyle b}}$

All numbers are in base ${\displaystyle{\displaystyle b}}$.

Base ${\displaystyle{\displaystyle b}}$	Power ${\displaystyle{\displaystyle p}}$	Nontrivial Kaprekar numbers ${\displaystyle{\displaystyle n\neq 0}}$, ${\displaystyle{\displaystyle n\neq 1}}$	Cycles
2	1	10	${\displaystyle{\displaystyle\varnothing}}$
3	1	2, 10	${\displaystyle{\displaystyle\varnothing}}$
4	1	3, 10	${\displaystyle{\displaystyle\varnothing}}$
5	1	4, 5, 10	${\displaystyle{\displaystyle\varnothing}}$
6	1	5, 6, 10	${\displaystyle{\displaystyle\varnothing}}$
7	1	3, 4, 6, 10	${\displaystyle{\displaystyle\varnothing}}$
8	1	7, 10	2 → 4 → 2
9	1	8, 10	${\displaystyle{\displaystyle\varnothing}}$
10	1	9, 10	${\displaystyle{\displaystyle\varnothing}}$
11	1	5, 6, A, 10	${\displaystyle{\displaystyle\varnothing}}$
12	1	B, 10	${\displaystyle{\displaystyle\varnothing}}$
13	1	4, 9, C, 10	${\displaystyle{\displaystyle\varnothing}}$
14	1	D, 10	${\displaystyle{\displaystyle\varnothing}}$
15	1	7, 8, E, 10	2 → 4 → 2 9 → B → 9
16	1	6, A, F, 10	${\displaystyle{\displaystyle\varnothing}}$
2	2	11	${\displaystyle{\displaystyle\varnothing}}$
3	2	22, 100	${\displaystyle{\displaystyle\varnothing}}$
4	2	12, 22, 33, 100	${\displaystyle{\displaystyle\varnothing}}$
5	2	14, 31, 44, 100	${\displaystyle{\displaystyle\varnothing}}$
6	2	23, 33, 55, 100	15 → 24 → 15 41 → 50 → 41
7	2	22, 45, 66, 100	${\displaystyle{\displaystyle\varnothing}}$
8	2	34, 44, 77, 100	4 → 20 → 4 11 → 22 → 11 45 → 56 → 45
2	3	111, 1000	10 → 100 → 10
3	3	111, 112, 222, 1000	10 → 100 → 10
2	4	110, 1010, 1111, 10000	${\displaystyle{\displaystyle\varnothing}}$
3	4	121, 2102, 2222, 10000	${\displaystyle{\displaystyle\varnothing}}$
2	5	11111, 100000	10 → 100 → 10000 → 1000 → 10 111 → 10010 → 1110 → 1010 → 111
3	5	11111, 22222, 100000	10 → 100 → 10000 → 1000 → 10
2	6	11100, 100100, 111111, 1000000	100 → 10000 → 100 1001 → 10010 → 1001 100101 → 101110 → 100101
3	6	10220, 20021, 101010, 121220, 202202, 212010, 222222, 1000000	100 → 10000 → 100 122012 → 201212 → 122012
2	7	1111111, 10000000	10 → 100 → 10000 → 10 1000 → 1000000 → 100000 → 1000 100110 → 101111 → 110010 → 1010111 → 1001100 → 111101 → 100110
3	7	1111111, 1111112, 2222222, 10000000	10 → 100 → 10000 → 10 1000 → 1000000 → 100000 → 1000 1111121 → 1111211 → 1121111 → 1111121
2	8	1010101, 1111000, 10001000, 10101011, 11001101, 11111111, 100000000	${\displaystyle{\displaystyle\varnothing}}$
3	8	2012021, 10121020, 12101210, 21121001, 20210202, 22222222, 100000000	${\displaystyle{\displaystyle\varnothing}}$
2	9	10010011, 101101101, 111111111, 1000000000	10 → 100 → 10000 → 100000000 → 10000000 → 100000 → 10 1000 → 1000000 → 1000 10011010 → 11010010 → 10011010

 امتداد للأعداد الصحيحة السالبة

Kaprekar numbers can be extended to the negative integers by use of a signed-digit representation to represent each integer.

 تمرين في البرمجة

The example below implements the Kaprekar function described in the definition above to search for Kaprekar numbers and cycles in Python.

def kaprekarf(x: int, p: int, b: int) -> int:
    beta = pow(x, 2) % pow(b, p)
    alpha = (pow(x, 2) - beta) // pow(b, p)
    y = alpha + beta
    return y

def kaprekarf_cycle(x: int, p: int, b: int) -> List[int]:
    seen = []
    while x < pow(b, p) and x not in seen:
        seen.append(x)
        x = kaprekarf(x, p, b)
    if x > pow(b, p):
        return []
    cycle = []
    while x not in cycle:
        cycle.append(x)
        x = kaprekarf(x, p, b)
    return cycle

 انظر أيضاً

• Arithmetic dynamics
• Automorphic number
• Dudeney number
• Factorion
• Happy number
• Kaprekar's constant
• Meertens number
• Narcissistic number
• Perfect digit-to-digit invariant
• Perfect digital invariant
• Sum-product number

 المراجع

• D. R. Kaprekar (1980–1981). "On Kaprekar numbers". Journal of Recreational Mathematics. 13: 81–82.
• M. Charosh (1981–1982). "Some Applications of Casting Out 999...'s". Journal of Recreational Mathematics. 14: 111–118.
• Iannucci, Douglas E. (2000). "The Kaprekar Numbers". Journal of Integer Sequences. 3: 00.1.2.