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I formulated this conjecture after reading this related question.

Let $\mathcal{P}(n) = \{P_1(n), P_2(n), \ldots \}$ be the set of all integer partitions of a positive integer $n$, and $p(n)=\vert \mathcal{P}(n) \vert$ the number of those partitions. Note that the $P_k(n)$ are multisets, i.e. elements can be repeated.

Let $f(P_k(n))$ be the number of distinct permutations of the elements of $P_k(n)$.

Is it true that for $n \ge 2$:

$$\sum_{k=1}^{p(n)} f(P_k(n))(-1)^{\vert P_k(n) \vert} = 0 \space ?$$

I verified it for $2 \le n \le 6$, while it evaluates to $-1$ for $n = 1$.

EDIT 1

Searching the OEIS and precisely the comments section of OEIS A111786 it seems this is a well known fact. I haven't found the proof yet...

EDIT 2

For example, for $n=4$, $\mathcal{P}(4) = \{\{4\},\{3,1\},\{2,2\},\{2,1,1\},\{1,1,1,1\}\}$, $f(\{4\})(-1)=-1$, $f(\{3,1\})(-1)^2=2$, $f(\{2,2\})(-1)^2=1$, $f(\{2,1,1\})(-1)^3=-3$, $f(\{1,1,1,1\})(-1)^4=1$.

Fabius Wiesner
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    It would help if you gave a few examples for some n you have checked, so users here can see better what your conjecture means in some concrete cases. – coffeemath Dec 07 '23 at 11:37
  • Using other language, $f(P_k(n))$ counts the number of integer compositions of $n$ that correspond to the partition $P_k(n)$. – Brian Hopkins Dec 09 '23 at 23:38

1 Answers1

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This is true and essentially boils down to the identity: $$ \sum_{|P_k|=m}f(P_k(n))=\binom {n-1}{m-1}. $$

The identity can be easily proved by stars and bars as the LHS represents the number of ways to distribute $n$ balls between $m$ bins provided that no bin is empty.

user
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