22
votes

Accepted

### an identity for a sum over partitions

Multiply the whole equality by $(-1)^n$.
First of all, notice that $k\choose m_1,\dots,m_k$ is the number of ways to permute the numbers $\lambda_1,\dots,\lambda_k$, so the left-hand side equals
$$
...

22
votes

Accepted

### powered partition function generator: 1/2 of them are zeros?

This is true. First, note that by the Pentagonal number theorem due to Euler,
$$\frac{1}{F(x)} = \sum_{k \in \mathbb{Z}} (-1)^k x^{\frac{k}{2}(3k-1)}.$$
For a given prime $p \ge 5$, the function $f(...

22
votes

Accepted

### A property of 47 with respect to partitions into five parts

Yes. Suppose $n>47$.
If $2\mid n$, we can take $(n-8,2,2,2,2),(n-10,4,2,2,2)$, which are distinct partitions for $n\geq 14$.
If $3\mid n$, we can take $(n-12,3,3,3,3),(n-15,6,3,3,3)$, which are ...

21
votes

Accepted

### Alternating sum of hook lengths: Part I

The hook length $h_{ij}(\lambda)$ counts the number of boxes directly below or directly to the right of box $(i,j)$. (I picture the Young diagram of $\lambda$ as having the corner $(0,0)$ located in ...

20
votes

### Does Rademacher's convergent series for p(n) define an analytic function?

Not a direct answer to the question, but a brief numerical exploration of this function.
First, a trivial observation: we can write either $e^{\pi i x}$ or $\cos(\pi x)$ in the formula for the ...

20
votes

### Bijective proof for a partition identity

The answer is yes, there is a combinatorial proof, and both Sam's and Fёdor's proofs work. However, this is a really old result and a combinatorial proof is old. Here is the reference:
H. Gupta, ...

19
votes

Accepted

### Does Rademacher's convergent series for p(n) define an analytic function?

Edit. We can write the series in the form
$$p(n)=\sum A_k(z)\frac{d}{dz}f(z/k^2),$$
where $|A_k(z)|\leq Ck^{1/2}e^{C_1(\Im z)^+},$ where $y^+=\max\{ y,0\},$ and $C_j$ are various positive absolute ...

18
votes

Accepted

### Inequality for hook numbers in Young diagrams

Not sure, please check carefully. (Well, now more sure and the argument is more direct.)
I claim that the array $(h)$ majorates the array $(q)$, that is,
$\sum \varphi (h_{ij})\geqslant \sum \varphi(...

18
votes

Accepted

### Two interpretations of a sequence: an opportunity for combinatorics

Here is a bijective proof that equates both $a(n)$ and $b(n+1)$ to the quantity
$$p(n)+2p(n-1)+\cdots+np(1)+(n+1)p(0) \tag1$$
where $p(n)$ is the number of partitions of $n$.
For $a(n)$: each ...

18
votes

Accepted

### Equality of two $q$-series. Proof?

I take it from looking at the previous problem that you are familiar with the Dyson rank on partitions with distinct parts. Let's denote by $Q(r,n)$ the number of partitions of $n$ into distinct parts ...

16
votes

Accepted

### In search of a combinatorial reasoning for a vanishing sum

For fixed $s,j>0$, $\sum_{\mathcal{A}_{j,s}}\binom{s}{n_1,\dots,n_j}$ enumerates the compositions of $j$ into $s$ positive parts by ordering the corresponding partitions. That is, we have
$$\sum_{\...

16
votes

Accepted

### Which of these sums appear most often?

In fact this question was already asked at MO, although in disguise: see here. Richard Stanley answered it wonderfully. The champions are the nearest integers to $n(n+1)/4$.
For a quick proof, see ...

15
votes

Accepted

### Identity involving a sum over all partitions of $n$

Here's a quick sketch (since I'm pressed for time). Multiply both sides of the identity by $t^n$ and sum over $n$ from $0$ to infinity. From the cycle decomposition identity (Polya's formula) the ...

15
votes

Accepted

### Are the Fourier coefficients of $\eta(q^m)^m / \eta(q)$ non-negative?

Yes this is true and these Fourier coefficients actually enumerate some combinatorial objects called $m$-cores. These are partitions with no hooklength divisible by $m$. In fact for $m\geq 4$ these ...

15
votes

Accepted

### Sum of squares and partitions

Start by checking that the following formal product can be expanded as a sum over partitions
$$\prod_{i\geq 1}\left(1+\sum_{r\geq 1}a_r(x_1x_2\cdots x_i)^r\right)=\sum_{\lambda}\left(\prod_{j\geq 1}a_{...

15
votes

Accepted

### Terminology for a bijection from a set to itself

Permutation is the term I would use (indeed, when I teach, I define a "permutation" of a set $X$ as a bijection from $X$ to itself).

14
votes

### an identity for a sum over partitions

Here is a purely combinatorial proof of a more general identity
$$
\sum_{\lambda\vdash n}(-1)^{n-k}\frac{k!}{m_1!\cdots m_n!}\prod_{i=1}^k\binom{A}{\lambda_i}=\binom{A+n-1}n.
$$
As Ilya suggests, we ...

14
votes

Accepted

### Congruences Ramanujan-style

More general versions of this have been established: see in particular Theorem 2 of Kiming and Olsson, and for other work see (for example) Locus and Wagner.
To answer the question fully, as Ofir ...

14
votes

### A generalization of partition function to the sums of squares

This asymptotic was stated by Hardy and Ramanujan, but without proof. The first proof of this asymptotic was given by Wright in 1934 [1], by a rather complicated argument. A much simpler approach ...

14
votes

### Number of d-Calabi-Yau partitions

When $d=n-2$ (which constitutes the bulk of the cases) these partitions are known as Egyptian fractions, and their number is tabulated at http://oeis.org/A002966. No closed formula seems to exist.
For ...

14
votes

Accepted

### Bijective proof for a partition identity

Lemma. For $n>1$, the number of partitions of $n$ onto an even number of powers of 2 (here powers of 2 are 1,2,4,...) and the number of partitions of $n$ onto an odd number of powers of 2 are equal....

13
votes

Accepted

### hooks and contents: Part I

Specialize the Cycle Index Formula to get
$$\prod_{n=1}^\infty \exp \bigl( \frac{a_i}{i}z^i \bigr) = \sum_{n=0}^\infty \frac{z^n}{n!} \sum_{\pi \in \mathfrak{S}_n} a_1^{\mathrm{cyc}_1(\pi)} a_2^{\...

13
votes

Accepted

### "strange" diophantine and parity of the partition function

Your conjecture is true! Here is one way to get it using some mod 4 generatingfunctionology. The answer got a bit long, so I divided it into two parts, as an attempt to improve readability.
Part1: ...

13
votes

Accepted

### Generating function for $3$-core partitions

The set of $3$-core partitions can be described explicitly.
Theorem The partition $\lambda=\{\lambda_1,\lambda_2,\dots\}$ of length $k$ (that is, $\lambda_k > 0$ but $\lambda_{k+1} = \lambda_{k+2}...

13
votes

### How many ways can $N$ be written as a sum of terms in the form $2^i3^j$?

The generating function is
$$\prod_{i \ge 0}\prod_{j \ge 0} \left(1+z^{2^i 3^j}\right),$$
which, by uniqueness of binary expansion, simplifies to
$$\prod_{k \ge 0} \frac{1}{1-z^{3^k}},$$
the ...

13
votes

Accepted

### Generating function for counting partitions with corners

Maybe this is not the kind of answer you are looking for, but it is easy to see that
$$ P(x,t) = \prod_{k=1}^{\infty} \frac{1-x^{k-1}(1-t)}{1-x^k}$$
because the number of corners of a partition is one ...

12
votes

### is this a familiar gen. fn. for partitions?

We have the identity
$$\frac{1}{1 - x^k} = \prod_{i \ge 0} (1 + x^{k \cdot 2^i})$$
which is equivalent to the uniqueness of binary representations, and is also straightforward to prove using a ...

11
votes

### An identity related to partitions into $n$ parts and Schur polynomials

We can use the fact that $N(\lambda)=\left|\text{SSYT}(\lambda)\right|$, the number of semistandard Young tableaux of shape $\lambda$, and that $d!\cdot\left(\frac{\prod_{1\le i < j\le n}(\lambda_i ...

11
votes

Accepted

### A binary hook-length formula?

I am afraid they are not always integers. Take large $p$ and $n=2^{2p}-1$. Then $[n]!_b$ is divisible by $p^N$ for $N={2p\choose p}+1$. And $[2n+1]!_b$ by $p^K$ for $K={2p+1\choose p}+2p+1<2N$. ...

11
votes

### Bijective proof for a partition identity

There's a well-known bijective proof that the number of self-conjugate partitions of $n$ is the same as the number of partitions of $n$ into distinct odd parts (see https://en.wikipedia.org/wiki/...

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