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I have a question about restricted partitions of numbers:

For $n$ and $k$ positive integers let $M$ be the multiset in which each positive integer less than n appears exactly $k$ times. I want to understand the partitions obtained from elements of $M$ of numbers $s$ which are equivalent to $0$ mod $n$.

With $n$ and $k$ given, let $F(s,v)$ be the number of ways to achieve $s$ as the sum of $v$ many elements of $M$. To be clear, if $k=2$ then $F(1,1)=2$. That is, different elements of $M$ with the same numerical value are counted separately.

My question is this: For $n$ fixed greater than 1, are there arbitrarily large $k$ such that

$\sum$($F(s,v)$ such that $s$ is equivalent to $0$ mod $n$ and $v$ is even)
greater than
$\sum$($F(s,v)$ such that $s$ is equivalent to $0$ mod $n$ and $v$ is odd)?

Also, is there much known about $F$ such as a closed expression or generating function? I would appreciate any pointers to references in the literature.

Thanks in advance.

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  • $\begingroup$ Some clarity is desired. I see at least two interpretations, where s is fixed and the question is asked for each such s, and where s is not fixed, all the odd part partitions are gathered in one multiset and all the even part ones in another multiset, and you ask which multiset is bigger. Also, are the elements of M labeled so that you care which occurrence of 3 appears in a partition? Gerhard "Ask Me About Being Confused" Paseman, 2013.06.13 $\endgroup$ Commented Jun 14, 2013 at 17:27
  • $\begingroup$ Also, if n is even and k is odd, I think an odd partition for s is the complement of an even partition for an s', so for a reasonable (mis?)interpretation of your question, the answer is that there are the same number of odd partitions as there are even partitions. Gerhard "Saw Someone Else In Mirror" Paseman, 2013.06.14 $\endgroup$ Commented Jun 14, 2013 at 17:33
  • $\begingroup$ I edited the question to make it more mathematical and, I hope, unambiguous. Thanks for being interested. $\endgroup$ Commented Jun 14, 2013 at 18:37
  • $\begingroup$ I would suggest F(s,1)=2 (or 0) in your clarifying example, and for purposes of generating functions, things might be more cleanly stated by using F(tn,v) for integral values of t instead of F(s,v) inside the sum. Except for when n and k are both odd, I don't know the answer. Gerhard "Ask Me About System Design" Paseman, 2013.06.14 $\endgroup$ Commented Jun 14, 2013 at 19:27
  • $\begingroup$ Oops, I think I mean n is even and k is odd. Gerhard "Is Sorry About That, Chief" Paseman, 2013.06.14 $\endgroup$ Commented Jun 14, 2013 at 19:31

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