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The generating function $g(x):=(1−k^2 x)^{-1/k}$ satisfies, besides $g(0)=1,$

$ g^k= 1 + k^2 x\ g^k $,

whence we may express $c(k,n)$ as a sum of products of $c(k,j)$, with $j < n$, showing inductively that they are all integers, and in fact, multiples of $k$ for $n>0$.

Indeed, hiding the variable $k$ in $c(k,n)=c(n)$, one has

$c(0)=1$, $c(1)=k$

and in general for n>1,

$c(n)= k\sum_\mu c(\mu_1)c(\mu_2)..c(\mu_k) - \frac{1}{k}\sum_\nu c(\nu_1)c(\nu_2)..c(\nu_k)$

the first sum being extended over all multi-indices $\mu\in \mathbb{N}^k$ with weight $|\mu|:=\mu_1+\mu_2\dots +\mu_k=n-1$, while the second over all $\nu\in \mathbb{N}^k$ with $|\nu|=n$ and $\nu_j< n$ for $j=1\dots k$. (The factor 1/k doesn't bother, because each term in the second sum contains at least 2 factors $c(j)$ with $0 < j < n$, and these by inductive hypothesis are multiples of $k$).

The generating function $g(x):=(1−k^2 x)^{-1/k}$ satisfies, besides $g(0)=1,$

$ g^k= 1 + k^2 x\ g^k $,

whence we may express $c(k,n)$ as a sum of products of $c(k,j)$, with $j < n$, showing inductively that they are all integers, and in fact, multiples of $k$ for $n>0$.

Indeed, hiding the variable $k$ in $c(k,n)=c(n)$, one has

$c(0)=1$, $c(1)=k$

and in general for n>1,

$c(n)= k\sum_\mu c(\mu_1)c(\mu_2)..c(\mu_k) - \frac{1}{k}\sum_\nu c(\nu_1)c(\nu_2)..c(\nu_k)$

the first sum being extended over all multi-indices $\mu\in \mathbb{N}^k$ with weight $|\mu|:=\mu_1+\mu_2\dots +\mu_k=n-1$, while the second over all $\nu\in \mathbb{N}^k$ with $|\nu|=n$ and $\nu_j< n$ for $j=1\dots k$. It follows that if $c(j)$ are multiple of $k$ for $1 < j < n$, so is $c(n)$, and by induction this proves the claim. (The factor 1/k doesn't bother, because each term in the second sum contains at least 2 factors $c(j)$ with $0 < j < n$).

The generating function $g(x):=(1−k^2 x)^{-1/k}$ satisfies, besides $g(0)=1,$

$ g^k= 1 + k^2 x\ g^k $,

whence we may express $c(k,n)$ as a certain linear combination of k-fold products of some $c(k,j)$, with $j < n$, showing inductively that they are all integers, and in fact, multiples of $k$ for $n>0$.

Precisely, hiding the variable $k$ in $c(k,n)=c(n)$, one has

$c(0)=1$, $c(1)=k$

and in general for n>1,

$c(n)= k\sum_\mu c(\mu_1)c(\mu_2)..c(\mu_k) - \frac{1}{k}\sum_\nu c(\nu_1)c(\nu_2)..c(\nu_k)$

the first sum being extended over all multiindices $\mu\in \mathbb{N}^k$ with weight $|\mu|:=\mu_1+\mu_2\dots +\mu_k=n-1$, while the second over all $\nu\in \mathbb{N}^k$ with $|\nu|=n$ and $\nu_j< n$ for $j=1\dots k$. (The factor 1/k doesn't bother, because each term in the second sum contains at least 2 factors $c(j)$ with $0 < j < n$, and these by inductive hypothesis are multiples of $k$).

The generating function $g(x):=(1−k^2 x)^{-1/k}$ satisfies, besides $g(0)=1,$

$ g^k= 1 + k^2 x\ g^k $,

whence we may express $c(k,n)$ as a sum of products of $c(k,j)$, with $j < n$, showing inductively that they are all integers, and in fact, multiples of $k$ for $n>0$.

Indeed, hiding the variable $k$ in $c(k,n)=c(n)$, one has

$c(0)=1$, $c(1)=k$

and in general for n>1,

$c(n)= k\sum_\mu c(\mu_1)c(\mu_2)..c(\mu_k) - \frac{1}{k}\sum_\nu c(\nu_1)c(\nu_2)..c(\nu_k)$

the first sum being extended over all multi-indices $\mu\in \mathbb{N}^k$ with weight $|\mu|:=\mu_1+\mu_2\dots +\mu_k=n-1$, while the second over all $\nu\in \mathbb{N}^k$ with $|\nu|=n$ and $\nu_j< n$ for $j=1\dots k$. (The factor 1/k doesn't bother, because each term in the second sum contains at least 2 factors $c(j)$ with $0 < j < n$, and these by inductive hypothesis are multiples of $k$).

The generating function $g(x):=(1−k^2 x)^{-1/k}$ satisfies, besides $g(0)=1,$

$ g^k= 1 + k^2 x\ g^k $,

whence we may express $c(k,n)$ as a certain linear combination of k-fold products of some $c(k,j)$, with $j < n$, showing inductively that they are all integers, and in fact, multiples of $k$ for $n>0$.

Precisely, hiding the variable $k$ in $c(k,n)=c(n)$, one has

$c(0)=1$, $c(1)=k$

and in general for n>0

$c(n)= k\sum_\mu c(\mu_1)c(\mu_2)..c(\mu_k) - \frac{1}{k}\sum_\nu c(\nu_1)c(\nu_2)..c(\nu_k)$

the first sum being extended over all multiindices $\mu\in \mathbb{N}^k$ with weight $|\mu|:=\mu_1+\mu_2\dots +\mu_k=n-1$, while the second over all $\nu\in \mathbb{N}^k$ with $|\nu|=n$ and $\nu_j< n$ for $j=1\dots k$.

The generating function $g(x):=(1−k^2 x)^{-1/k}$ satisfies, besides $g(0)=1,$

$ g^k= 1 + k^2 x\ g^k $,

whence we may express $c(k,n)$ as a certain linear combination of k-fold products of some $c(k,j)$, with $j < n$, showing inductively that they are all integers, and in fact, multiples of $k$ for $n>0$.

Precisely, hiding the variable $k$ in $c(k,n)=c(n)$, one has

$c(0)=1$, $c(1)=k$

and in general for n>1,

$c(n)= k\sum_\mu c(\mu_1)c(\mu_2)..c(\mu_k) - \frac{1}{k}\sum_\nu c(\nu_1)c(\nu_2)..c(\nu_k)$

the first sum being extended over all multiindices $\mu\in \mathbb{N}^k$ with weight $|\mu|:=\mu_1+\mu_2\dots +\mu_k=n-1$, while the second over all $\nu\in \mathbb{N}^k$ with $|\nu|=n$ and $\nu_j< n$ for $j=1\dots k$. (The factor 1/k doesn't bother, because each term in the second sum contains at least 2 factors $c(j)$ with $0 < j < n$, and these by inductive hypothesis are multiples of $k$).