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27 votes
3 answers
2k views

Kasteleyn's formula for domino tilings generalized?

It seems a marvel when a bunch of irrational numbers "conspire" to become rational, even better an integer. An elementary example is $\prod_{j=1}^n4\cos^2\left(\pi j/(2n+1)\right)=1$. Kasteleyn's ...
T. Amdeberhan's user avatar
23 votes
4 answers
2k views

Identity for an infinite product

Here is an experimental "result" exhibiting the difference of two (formal) infinite products that "almost factorizes". QUESTION. Is this true? $$\prod_{n\geq1}(1+x^{2n-1})^{24} - \...
T. Amdeberhan's user avatar
16 votes
6 answers
2k views

Alternative proofs sought after for a certain identity

Here is an identity for which I outlined two different arguments. I'm collecting further alternative proofs, so QUESTION. can you provide another verification for the problem below? Problem. Prove ...
T. Amdeberhan's user avatar
12 votes
1 answer
596 views

Equality of two $q$-series. Proof?

Recall the notation $(z;q)_n=(1-z)(1-zq)(1-zq^2)\cdots(1-zq^{n-1})$. My earlier MO question did not find enough interest or yield an answer. Perhaps the modulo $2$ part might have thrown people off. ...
T. Amdeberhan's user avatar
11 votes
2 answers
539 views

Reference request: A multidimensional generalization of the fundamental theorem of calculus

$\newcommand\R{\mathbb R}$Let $f\colon\R^p\to\R$ be a continuous function. For $u=(u_1,\dots,u_p)$ and $v=(v_1,\dots,v_p)$ in $\R^p$, let $[u,v]:=\prod_{r=1}^p[u_r,v_r]$; $u\wedge v:=\big(\min(u_1,v_1)...
Iosif Pinelis's user avatar
10 votes
0 answers
287 views

Coefficients of polynomials vs trigonometric product

Let's consider the family of sequences of coefficients in the expansion $$\prod_{i=0}^{n-1}(1+x^{3^i}+x^{3^{i+1}})=\sum_{k\geq0}a_n(k)\, x^k.$$ Remark. Evidently, the RHS is a finite sum. Here is a ...
T. Amdeberhan's user avatar
9 votes
2 answers
354 views

Asymptotics of a quadratic recursion

Consider the sequence defined by \begin{align} c_0 &{}= 1 \\ c_n &{}= 2\,n\,c_{n-1}-\frac{1}{2}\sum_{m=1}^{n-1}c_m\,c_{n-m}. \end{align} How can you prove that it has the following asymptotics ...
Matteo Beccaria's user avatar
6 votes
2 answers
755 views

Prove positivity of a binomial sum

Some problems appear easy on the face of it, but perhaps they are not. Here is an instance of a certain calculation which is slightly reformulated from its original encounter in a current work. I have ...
T. Amdeberhan's user avatar
5 votes
0 answers
107 views

Generalized Puiseux series for diagonal reflections of the curves $y = \frac{x}{(1-ax)(1-bx)^m}$

Reflection of the curve $y = f_m(x) = \frac{x}{(1-ax)(1-bx)^m}$ through the diagonal line $y=x$ in the $xy$-plane can be regarded as local compositional inversion of the curve $y=f_m(x)$. ($x,y,a,b$ ...
Tom Copeland's user avatar
  • 10.5k
4 votes
1 answer
246 views

Is $C_n$ infinitely log-convex?

A sequence $a_n$ is called log-convex if $\mathcal{L}(a_n):=a_{n+1}a_{n-1}-a_n^2\geq0$ for all $n$; it is infinitely log-convex provided that all the iterates $\mathcal{L}^k(a_n)$ are still log-convex,...
T. Amdeberhan's user avatar
4 votes
1 answer
95 views

Limiting values of particular functions

Let's define the functions $$A_n(q)=\sum_{k=0}^n(-1)^k\cdot\frac{(1+q)q^k}{1+q^{2k+1}}\cdot\frac{2k+1}{n+k+1}\binom{2n}{n-k}.$$ I'm interested in the following: QUESTION. Let $n\geq1$ be integers. ...
T. Amdeberhan's user avatar
4 votes
0 answers
208 views

Extract this constant term

Given a Laurent polynomial $F$ in the variables $\mathbf{t}=(t_1,\dots,t_n)$, let $CT_{\vec{\mathbf{t}}}\,F$ denote its constant term. For example, $CT_{t_1,t_2}((8t_1-\frac1{3t_1t_2})(5t_1t_2+t_2^2+\...
T. Amdeberhan's user avatar
3 votes
1 answer
251 views

Congruence modulo 2 for q-series

This quest arose from certain calculations with integer partitions (having distinct parts) and the corresponding values of their Dyson ranks. I would like to ask: QUESTION. Is this congruence true ...
T. Amdeberhan's user avatar
3 votes
0 answers
315 views

When does the Taylor coefficient of $e^{\sin x}$ vanish?

If $f(x)=\frac{a_1}{1!}x+\frac{a_2}{2!}x^2+\frac{a_3}{3!}x^3+\frac{a_4}{4!}x^4+\cdots$ is an exponential generating function for $\{a_k\}_{k\geq1}$ then $$e^{f(x)}=1+\frac{a_1}{1!}x+\frac{a_1^2+a_2}{2!...
T. Amdeberhan's user avatar
2 votes
1 answer
377 views

Prove positivity of rational functions

We say a rational function $F(z)$ is positive if the coefficients of its Maclaurin expansion, in the variable $z$, are non-negative. In this context, let $$F_r(z):=\frac{1 - 2z + z^r - (1 - z)^r}{(1 - ...
T. Amdeberhan's user avatar
2 votes
1 answer
276 views

Estimating a sum over set partitions

Let $[n]:=\{1,\dots,n\}$. Fix a set partition $\rho$ of $[n]$, with an abuse of notation we shall use $\rho\vdash [n]$. I would like to estimate the following alternating sum. QUESTION. Is this true? ...
T. Amdeberhan's user avatar
1 vote
1 answer
117 views

Product/quotient of factorials beget dyadic powers

I am writing up some notes and the following occurred to me and I would like to see if there are a variety of ways to prove it. Just for reference, the identity pops out of equality between constant ...
T. Amdeberhan's user avatar
1 vote
1 answer
457 views

A (surprising?) expression for $e$

I apologise if this is off topic. Consider the quantity $$ F(m,n,k)=\frac{(m)_k}{k!n^{k-1} } $$ where $m,n \in \mathbb{N}.$ For moderately large $n$, it seems that the approximation $$ \sum_{k=1}^{K} ...
kodlu's user avatar
  • 10.4k
1 vote
0 answers
162 views

Triangular and pentagonal numbers in $q$-series

Consider the following two infinite series $$\sum_{n\geq0}a(n)q^n=\prod_{k\geq1}\frac1{(1-q^k)^2(1-q^{5k})^2} \,\,\,\, \text{and} \,\,\, \sum_{n\geq0}b(n)q^n=\prod_{k\geq1}\frac1{(1-q^k)^2(1-q^{7k})^2}...
T. Amdeberhan's user avatar
0 votes
1 answer
129 views

Seeking an integral formulation for an algebraic function

While working with a generating function for the Catalan numbers, I came across the integral representation $$\frac1{1+\sqrt{1-4x}}=\frac1{2\pi}\int_0^{\infty}\frac{\sqrt{t}}{(t+\frac14)(t-x+\frac14)}\...
T. Amdeberhan's user avatar