Kaprekar number

Base-dependent property of integers

In mathematics, a natural number in a given number base is a p {\displaystyle p} -Kaprekar number if the representation of its square in that base can be split into two parts, where the second part has p {\displaystyle p} digits, that add up to the original number. For example, in base 10, 45 is a 2-Kaprekar number, because 45² = 2025, and 20 + 25 = 45. The numbers are named after D. R. Kaprekar.

Definition and properties

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

F p , b ( n ) = α + β {\displaystyle F_{p,b}(n)=\alpha +\beta } ,

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

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

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

The earlier example of 45 satisfies this definition with b = 10 {\displaystyle b=10} and p = 2 {\displaystyle p=2} , because

β = n 2 mod b p = 45 2 mod 1 0 2 = 25 {\displaystyle \beta =n^{2}{\bmod {b}}^{p}=45^{2}{\bmod {1}}0^{2}=25}
α = n 2 β b p = 45 2 25 10 2 = 20 {\displaystyle \alpha ={\frac {n^{2}-\beta }{b^{p}}}={\frac {45^{2}-25}{10^{2}}}=20}
F 2 , 10 ( 45 ) = α + β = 20 + 25 = 45 {\displaystyle F_{2,10}(45)=\alpha +\beta =20+25=45}

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

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

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

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

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

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

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

Set-theoretic definition and unitary divisors

We can define the set K ( N ) {\displaystyle K(N)} for a given integer N {\displaystyle N} as the set of integers X {\displaystyle X} for which there exist natural numbers A {\displaystyle A} and B {\displaystyle B} satisfying the Diophantine equation[1]

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

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

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

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

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

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

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

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

Let

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

Then,

X 1 2 = ( b p 1 2 ) 2 = b 2 p 2 b p + 1 4 = b p ( b p 2 ) + 1 4 = ( 4 k + 3 ) 2 n + 1 ( b p 2 ) + 1 4 = ( 4 k + 3 ) 2 n ( b p 2 ) ( 4 k + 4 ) ( 4 k + 3 ) 2 n ( b p 2 ) + 1 4 = ( 4 k + 3 ) 2 n ( b p 2 ) + 1 4 + ( k + 1 ) ( 4 k + 3 ) 2 n ( b p 2 ) = ( 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 ) = ( 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 ) = ( 4 k + 3 ) p 2 ( b p 2 ) + 1 4 + i = p 2 p 1 ( 1 ) i ( k + 1 ) b i ( b p 2 ) = ( 4 k + 3 ) p 2 ( b p 2 ) + 1 4 + ( b p 2 ) ( k + 1 ) i = p 2 p 1 ( 1 ) i b i = ( 4 k + 3 ) 1 ( b p 2 ) + 1 4 + ( b p 2 ) ( k + 1 ) i = 1 p 1 ( 1 ) i b i = ( b p 2 ) + 1 4 + ( b p 2 ) ( k + 1 ) i = 0 p 1 ( 1 ) i b i = ( b p 2 ) ( k + 1 ) ( i = 0 2 n ( 1 ) i b i ) + b 2 n + 1 + 3 4 = ( b p 2 ) ( k + 1 ) ( i = 0 2 n ( 1 ) i b i ) + 4 b 2 n + 1 + 3 b 2 n + 1 + 3 4 = ( b p 2 ) ( k + 1 ) ( i = 0 2 n ( 1 ) i b i ) b p + 3 b 2 n + 1 + 3 4 = ( b p 2 ) ( k + 1 ) ( i = 0 2 n ( 1 ) i b i ) b p + 3 ( 4 k + 3 ) p 2 + 3 4 + 3 ( k + 1 ) i = p 2 p 1 ( 1 ) i b i = ( b p 2 ) ( k + 1 ) ( i = 0 2 n ( 1 ) i b i ) b p + 3 ( 4 k + 3 ) 1 + 3 4 + 3 ( k + 1 ) i = 1 p 1 ( 1 ) i b i = ( b p 2 ) ( k + 1 ) ( i = 0 2 n ( 1 ) i b i ) b p + 3 + 3 4 + 3 ( k + 1 ) i = 0 p 1 ( 1 ) i b i = ( b p 2 ) ( k + 1 ) ( i = 0 2 n ( 1 ) i b i ) + 3 ( k + 1 ) ( i = 0 2 n ( 1 ) i b i ) b p = ( b p 2 + 3 ) ( k + 1 ) ( i = 0 2 n ( 1 ) i b i ) b p = ( b p + 1 ) ( k + 1 ) ( i = 0 2 n ( 1 ) i b i ) b p = ( b p + 1 ) ( 1 + ( k + 1 ) i = 0 2 n ( 1 ) i b i ) + 1 = ( b p + 1 ) ( k + ( k + 1 ) i = 1 2 n ( 1 ) i b i ) + 1 = ( b p + 1 ) ( k + ( k + 1 ) i = 1 n b 2 i b 2 i 1 ) + 1 = ( b p + 1 ) ( k + ( k + 1 ) i = 1 n ( b 1 ) b 2 i 1 ) + 1 = ( b p + 1 ) ( k + i = 1 n ( ( k + 1 ) b k 1 ) b 2 i 1 ) + 1 = ( b p + 1 ) ( k + i = 1 n ( k b + ( 4 k + 3 ) k 1 ) b 2 i 1 ) + 1 = ( b p + 1 ) ( k + i = 1 n ( k b + ( 3 k + 2 ) ) b 2 i 1 ) + 1 = b p ( k + i = 1 n ( k b + ( 3 k + 2 ) ) b 2 i 1 ) + ( k + 1 + i = 1 n ( k b + ( 3 k + 2 ) ) b 2 i 1 ) {\displaystyle {\begin{aligned}X_{1}^{2}&=\left({\frac {b^{p}-1}{2}}\right)^{2}\\&={\frac {b^{2p}-2b^{p}+1}{4}}\\&={\frac {b^{p}(b^{p}-2)+1}{4}}\\&={\frac {(4k+3)^{2n+1}(b^{p}-2)+1}{4}}\\&={\frac {(4k+3)^{2n}(b^{p}-2)(4k+4)-(4k+3)^{2n}(b^{p}-2)+1}{4}}\\&={\frac {-(4k+3)^{2n}(b^{p}-2)+1}{4}}+(k+1)(4k+3)^{2n}(b^{p}-2)\\&={\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)\\&={\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)\\&={\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)\\&={\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}\\&={\frac {(4k+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)\left(\sum _{i=0}^{2n}(-1)^{i}b^{i}\right)+{\frac {-b^{2n+1}+3}{4}}\\&=(b^{p}-2)(k+1)\left(\sum _{i=0}^{2n}(-1)^{i}b^{i}\right)+{\frac {-4b^{2n+1}+3b^{2n+1}+3}{4}}\\&=(b^{p}-2)(k+1)\left(\sum _{i=0}^{2n}(-1)^{i}b^{i}\right)-b^{p}+{\frac {3b^{2n+1}+3}{4}}\\&=(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}\\&=(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}\\&=(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}\\&=(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}\\&=(b^{p}-2+3)(k+1)\left(\sum _{i=0}^{2n}(-1)^{i}b^{i}\right)-b^{p}\\&=(b^{p}+1)(k+1)\left(\sum _{i=0}^{2n}(-1)^{i}b^{i}\right)-b^{p}\\&=(b^{p}+1)\left(-1+(k+1)\sum _{i=0}^{2n}(-1)^{i}b^{i}\right)+1\\&=(b^{p}+1)\left(k+(k+1)\sum _{i=1}^{2n}(-1)^{i}b^{i}\right)+1\\&=(b^{p}+1)\left(k+(k+1)\sum _{i=1}^{n}b^{2i}-b^{2i-1}\right)+1\\&=(b^{p}+1)\left(k+(k+1)\sum _{i=1}^{n}(b-1)b^{2i-1}\right)+1\\&=(b^{p}+1)\left(k+\sum _{i=1}^{n}((k+1)b-k-1)b^{2i-1}\right)+1\\&=(b^{p}+1)\left(k+\sum _{i=1}^{n}(kb+(4k+3)-k-1)b^{2i-1}\right)+1\\&=(b^{p}+1)\left(k+\sum _{i=1}^{n}(kb+(3k+2))b^{2i-1}\right)+1\\&=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 α {\displaystyle \alpha } and β {\displaystyle \beta } are

β = X 1 2 mod b p = k + 1 + i = 1 n ( k b + ( 3 k + 2 ) ) b 2 i 1 {\displaystyle \beta =X_{1}^{2}{\bmod {b}}^{p}=k+1+\sum _{i=1}^{n}(kb+(3k+2))b^{2i-1}}
α = X 1 2 β b p = k + i = 1 n ( k b + ( 3 k + 2 ) ) b 2 i 1 {\displaystyle \alpha ={\frac {X_{1}^{2}-\beta }{b^{p}}}=k+\sum _{i=1}^{n}(kb+(3k+2))b^{2i-1}}

and their sum is

α + β = ( k + i = 1 n ( k b + ( 3 k + 2 ) ) b 2 i 1 ) + ( k + 1 + i = 1 n ( k b + ( 3 k + 2 ) ) b 2 i 1 ) = 2 k + 1 + i = 1 n ( ( 2 k ) b + 2 ( 3 k + 2 ) ) b 2 i 1 = 2 k + 1 + i = 1 n ( ( 2 k ) b + ( 6 k + 4 ) ) b 2 i 1 = 2 k + 1 + i = 1 n ( ( 2 k ) b + ( 4 k + 3 ) ) b 2 i 1 + ( 2 k + 1 ) b 2 i 1 = 2 k + 1 + i = 1 n ( ( 2 k + 1 ) b ) b 2 i 1 + ( 2 k + 1 ) b 2 i 1 = 2 k + 1 + i = 1 n ( 2 k + 1 ) b 2 i + ( 2 k + 1 ) b 2 i 1 = 2 k + 1 + i = 1 2 n ( 2 k + 1 ) b i = i = 0 2 n ( 2 k + 1 ) b i = ( 2 k + 1 ) i = 0 2 n b i = X 1 {\displaystyle {\begin{aligned}\alpha +\beta &=\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)\\&=2k+1+\sum _{i=1}^{n}((2k)b+2(3k+2))b^{2i-1}\\&=2k+1+\sum _{i=1}^{n}((2k)b+(6k+4))b^{2i-1}\\&=2k+1+\sum _{i=1}^{n}((2k)b+(4k+3))b^{2i-1}+(2k+1)b^{2i-1}\\&=2k+1+\sum _{i=1}^{n}((2k+1)b)b^{2i-1}+(2k+1)b^{2i-1}\\&=2k+1+\sum _{i=1}^{n}(2k+1)b^{2i}+(2k+1)b^{2i-1}\\&=2k+1+\sum _{i=1}^{2n}(2k+1)b^{i}\\&=\sum _{i=0}^{2n}(2k+1)b^{i}\\&=(2k+1)\sum _{i=0}^{2n}b^{i}&=X_{1}\\\end{aligned}}}

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

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

Let

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

Then,

X 2 2 = ( X 1 + 1 ) 2 = X 1 2 + 2 X 1 + 1 = X 1 2 + 2 X 1 + 1 = b p ( k + i = 1 n ( k b + ( 3 k + 2 ) ) b 2 i 1 ) + ( k + 1 + i = 1 n ( k b + ( 3 k + 2 ) ) b 2 i 1 ) + b p 1 + 1 = b p ( k + 1 + i = 1 n ( k b + ( 3 k + 2 ) ) b 2 i 1 ) + ( k + 1 + i = 1 n ( k b + ( 3 k + 2 ) ) b 2 i 1 ) {\displaystyle {\begin{aligned}X_{2}^{2}&=(X_{1}+1)^{2}\\&=X_{1}^{2}+2X_{1}+1\\&=X_{1}^{2}+2X_{1}+1\\&=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\\&=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 α {\displaystyle \alpha } and β {\displaystyle \beta } are

β = X 2 2 mod b p = k + 1 + i = 1 n ( k b + ( 3 k + 2 ) ) b 2 i 1 {\displaystyle \beta =X_{2}^{2}{\bmod {b}}^{p}=k+1+\sum _{i=1}^{n}(kb+(3k+2))b^{2i-1}}
α = X 2 2 β b p = k + 1 + i = 1 n ( k b + ( 3 k + 2 ) ) b 2 i 1 {\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

α + β = ( k + 1 + i = 1 n ( k b + ( 3 k + 2 ) ) b 2 i 1 ) + ( k + 1 + i = 1 n ( k b + ( 3 k + 2 ) ) b 2 i 1 ) = 2 k + 2 + i = 1 n ( ( 2 k ) b + 2 ( 3 k + 2 ) ) b 2 i 1 = 2 k + 2 + i = 1 n ( ( 2 k ) b + ( 6 k + 4 ) ) b 2 i 1 = 2 k + 2 + i = 1 n ( ( 2 k ) b + ( 4 k + 3 ) ) b 2 i 1 + ( 2 k + 1 ) b 2 i 1 = 2 k + 2 + i = 1 n ( ( 2 k + 1 ) b ) b 2 i 1 + ( 2 k + 1 ) b 2 i 1 = 2 k + 2 + i = 1 n ( 2 k + 1 ) b 2 i + ( 2 k + 1 ) b 2 i 1 = 2 k + 2 + i = 1 2 n ( 2 k + 1 ) b i = 1 + i = 0 2 n ( 2 k + 1 ) b i = 1 + ( 2 k + 1 ) i = 0 2 n b i = 1 + X 1 = X 2 {\displaystyle {\begin{aligned}\alpha +\beta &=\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)\\&=2k+2+\sum _{i=1}^{n}((2k)b+2(3k+2))b^{2i-1}\\&=2k+2+\sum _{i=1}^{n}((2k)b+(6k+4))b^{2i-1}\\&=2k+2+\sum _{i=1}^{n}((2k)b+(4k+3))b^{2i-1}+(2k+1)b^{2i-1}\\&=2k+2+\sum _{i=1}^{n}((2k+1)b)b^{2i-1}+(2k+1)b^{2i-1}\\&=2k+2+\sum _{i=1}^{n}(2k+1)b^{2i}+(2k+1)b^{2i-1}\\&=2k+2+\sum _{i=1}^{2n}(2k+1)b^{i}\\&=1+\sum _{i=0}^{2n}(2k+1)b^{i}\\&=1+(2k+1)\sum _{i=0}^{2n}b^{i}\\&=1+X_{1}\\&=X_{2}\end{aligned}}}

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

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

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

  • X 1 = b p 1 m = ( m k + 1 ) i = 0 p 1 b i {\displaystyle X_{1}={\frac {b^{p}-1}{m}}=(mk+1)\sum _{i=0}^{p-1}b^{i}} is a Kaprekar number.
  • X 2 = b p + m 1 m = X 1 + 1 {\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 m {\displaystyle m} , k {\displaystyle k} , and n {\displaystyle n} be natural numbers, the number base b = m 2 k + m + 1 {\displaystyle b=m^{2}k+m+1} , and the power p = m n + m 1 {\displaystyle p=mn+m-1} . Then:

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

b = m2k + m2m + 1 and p = mn + 1

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

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

b = m2k + m2m + 1 and p = mn + m − 1

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

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

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

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

Base b {\displaystyle b} Power p {\displaystyle p} Nontrivial Kaprekar numbers n 0 {\displaystyle n\neq 0} , n 1 {\displaystyle n\neq 1} Cycles
2 1 10 {\displaystyle \varnothing }
3 1 2, 10 {\displaystyle \varnothing }
4 1 3, 10 {\displaystyle \varnothing }
5 1 4, 5, 10 {\displaystyle \varnothing }
6 1 5, 6, 10 {\displaystyle \varnothing }
7 1 3, 4, 6, 10 {\displaystyle \varnothing }
8 1 7, 10 2 → 4 → 2
9 1 8, 10 {\displaystyle \varnothing }
10 1 9, 10 {\displaystyle \varnothing }
11 1 5, 6, A, 10 {\displaystyle \varnothing }
12 1 B, 10 {\displaystyle \varnothing }
13 1 4, 9, C, 10 {\displaystyle \varnothing }
14 1 D, 10 {\displaystyle \varnothing }
15 1 7, 8, E, 10

2 → 4 → 2

9 → B → 9

16 1 6, A, F, 10 {\displaystyle \varnothing }
2 2 11 {\displaystyle \varnothing }
3 2 22, 100 {\displaystyle \varnothing }
4 2 12, 22, 33, 100 {\displaystyle \varnothing }
5 2 14, 31, 44, 100 {\displaystyle \varnothing }
6 2 23, 33, 55, 100

15 → 24 → 15

41 → 50 → 41

7 2 22, 45, 66, 100 {\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 \varnothing }
3 4 121, 2102, 2222, 10000 {\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 \varnothing }
3 8 2012021, 10121020, 12101210, 21121001, 20210202, 22222222, 100000000 {\displaystyle \varnothing }
2 9 10010011, 101101101, 111111111, 1000000000

10 → 100 → 10000 → 100000000 → 10000000 → 100000 → 10

1000 → 1000000 → 1000

10011010 → 11010010 → 10011010

Extension to negative integers

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

See also

Notes

  1. ^ a b Iannucci (2000)

References

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