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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 $$ ...
Ilya Bogdanov's user avatar
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(...
Ofir Gorodetsky's user avatar
22 votes
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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 ...
Wojowu's user avatar
  • 27.4k
21 votes
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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 ...
Gjergji Zaimi's user avatar
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 ...
Fredrik Johansson's user avatar
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, ...
Igor Pak's user avatar
  • 16.8k
19 votes
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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 ...
Alexandre Eremenko's user avatar
18 votes
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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(...
Fedor Petrov's user avatar
18 votes
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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 ...
Gjergji Zaimi's user avatar
18 votes
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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 ...
Gjergji Zaimi's user avatar
16 votes
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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_{\...
Max Alekseyev's user avatar
16 votes
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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 ...
GH from MO's user avatar
  • 101k
15 votes
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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 ...
Lucia's user avatar
  • 43.5k
15 votes
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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 ...
Gjergji Zaimi's user avatar
15 votes
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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_{...
Gjergji Zaimi's user avatar
15 votes
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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).
Max Horn's user avatar
  • 5,294
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 ...
Fedor Petrov's user avatar
14 votes
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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 ...
Lucia's user avatar
  • 43.5k
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 ...
Thomas Bloom's user avatar
  • 6,628
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 ...
pbelmans's user avatar
  • 1,496
14 votes
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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....
Fedor Petrov's user avatar
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^{\...
Mark Wildon's user avatar
  • 10.8k
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: ...
Gjergji Zaimi's user avatar
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}...
Gjergji Zaimi's user avatar
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 ...
RobPratt's user avatar
  • 5,179
13 votes
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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 ...
Sam Hopkins's user avatar
  • 23.1k
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 ...
Qiaochu Yuan's user avatar
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 ...
Gjergji Zaimi's user avatar
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 $K={2p+1\choose p}+2p+1<2N$. ...
Fedor Petrov's user avatar
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/...
Sam Hopkins's user avatar
  • 23.1k

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