All Questions
18 questions
13
votes
1
answer
468
views
Four new series for $\pi$ and related identities involving harmonic numbers
Recently, I discovered the following four new (conjectural) series for $\pi$:
\begin{align}\sum_{k=1}^\infty\frac{(5k^2-4k+1)8^k\binom{3k}k}{k(3k-1)(3k-2)\binom{2k}k\binom{4k}{2k}}&=\frac{3\pi}2,\...
- 15.6k
6
votes
0
answers
235
views
A curious series for $L(2,(\frac{-3}{\cdot}))$
Let
$$K:=L\left(2,\left(\frac{-3}{\cdot}\right)\right)=\sum_{k=1}^\infty\frac{(\frac k3)}{k^2}=\sum_{j=0}^\infty\left(\frac1{(3j+1)^2}-\frac1{(3j+2)^2}\right),$$ where $(\frac k3)$ is the Legendre ...
- 15.6k
6
votes
1
answer
484
views
Three conjectural series for $\pi^2$ and related identities
Recently, I found the following three (conjectural) identities for $\pi^2$:
$$\sum_{k=1}^\infty\frac{145k^2-104k+18}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{\pi^2}3,\tag{1}$$
$$\sum_{k=1}^\infty\frac{...
- 15.6k
6
votes
0
answers
298
views
A new series for $\sqrt3/\pi$?
Recently, I conjectured the following identity:
$$\sum_{k=0}^\infty\frac{(66k^2+37k+4)\binom{2k}k\binom{3k}k\binom{4k}{2k}}{(2k+1)729^k}=\frac{27\sqrt3}{2\pi}.\tag{1}$$
This can be easily checked ...
- 15.6k
3
votes
3
answers
756
views
Ordinary partitions vs partitions into odd parts
Let $\mathcal{P}(n)$ be the set of all unrestricted partitions of $n$ while $\mathcal{O}(n)$ stand for the set of all partitions of $n$ into odd parts. We adopt the power notation for partitions $\...
- 43.2k
4
votes
0
answers
97
views
"Convolving" a general Catalan with classical Catalan
Consider what is sometimes known as generalized Catalan sequence
$$\mathcal{{\color{red}C}}_{a,b}:=\frac{2b+1}{a+b+1}\binom{2a}{a+b}.$$
Observe that $\mathcal{{\color{red}C}}_{n,0}$ reduces to the ...
- 43.2k
1
vote
1
answer
220
views
Gaussian at $q=\pm1$, log-concave polynomials, Catalan numbers
Let $[n]_q!=\prod_{j=1}^n\frac{1-q^j}{1-q}$ with $[0]_q!:=1$ and the Gaussian polynomials $\binom{n}k_q=\frac{[n]_q!}{[k]_q!\,\cdot\,[n-k]_q!}$. Adopt the convention that $\binom{n}k_q=0$ whenever $k&...
- 43.2k
12
votes
3
answers
892
views
Set partitions and permanents
Let $a(n)=$ Number of ordered set partitions of $[n]$ such that the smallest element of each block is odd.
...
- 884
9
votes
1
answer
676
views
Permanent identities
The permanent $\mathrm{per}(A)$ of a matrix $A$ of size $n\times n$ is defined to be:
$$\mathrm{per}(A)=\sum_{\tau\in S_n}\prod_{j=1}^na_{j,\tau(j)}.$$
Let
$$A=\left[\tan\pi\frac{j+k}n\right]_{1\le j,...
- 884
3
votes
1
answer
439
views
An identity for polynomials over partitions
Given an integer partition $\lambda=(\lambda_1,\dots,\lambda_{\ell(\lambda)})$ of $n$ where $\ell(\lambda)$ is the length of $\lambda$, associate its conjugate partition $\lambda'$. Denote by $\lambda'...
- 43.2k
-1
votes
1
answer
271
views
How do I calculate this sum $\sum_k(k!)^{-n}$? [closed]
How do I evaluate the following finite sum over $k$
$1+\frac{1}{2^n}+\frac{1}{2^n3^n}+\frac{1}{2^n3^n4^n}+\cdots+\frac{1}{2^n3^n\cdots k^n}$
or if there is an expression of this sum in terms of ...
- 23
6
votes
2
answers
1k
views
Products and sum of cubes in Fibonacci
Consider the familiar sequence of Fibonacci numbers: $F_0=0, F_1=1, F_n=F_{n-1}+F_{n-2}$.
Although it is rather easy to furnish an algebraic verification of the below identity, I wish to see a ...
- 43.2k
24
votes
1
answer
593
views
Has the $E_8$-based generating function for squares numbers been proven?
In his 2004 paper Conformal Field Theory and Torsion Elements of the Bloch Group, Nahm explains a physical argument due to Kadem, Klassen, McCoy, and Melzer for the following remarkable identity. Let $...
- 54.6k
4
votes
3
answers
322
views
An identity for product of central binomials
This "innocent-looking" identity came out of some calculation with determinants, and I like to inquire if one can provide a proof. Actually, different methods of proofs would be of valuable merit and ...
- 43.2k
6
votes
2
answers
519
views
Seeking for a meaning: a curious symmetry
Suppose $\Phi(m,n)=(2m)!^n\prod_{k=1}^n\binom{2m+2k+x}{2k+x}$.
Then, algebraically, it is trivial to see that
$$(2m)!^n\prod_{k=1}^n\binom{2m+2k+x}{2k+x}=(2n)!^m\prod_{k=1}^m\binom{2n+2k+x}{2k+x}.$$
...
- 43.2k
8
votes
1
answer
472
views
In search of a combinatorial reasoning for a vanishing sum
Assume $s, j \in\mathbb{N}$. Define the set
$$\mathcal{A}_{j,s}:=\{(n_1,n_2,\dots,n_j)\in\mathbb{Z}_{\geq0}^j\vert \,
n_1+2n_2+\cdots+jn_j=j, \, n_1+n_2+\cdots+n_j=s\}.$$
Question. Is there a ...
- 43.2k
2
votes
1
answer
1k
views
A formula combining Euler $\phi$ and $\gcd$
Let us fix a natural number $N>1$ and $a_1, \ldots, a_n$ natural numbers satisfying $0 \leq a_i < N$, with the property that $1+ \sum a_i$ is divisible by $N$. Let $\phi$ be the Euler totient ...
- 283
2
votes
2
answers
894
views
Proving generating functions equality
What do you use to prove the following equality (and possibly more general ones of the kind)?
\begin{align*}\sum_{r,s,t} \frac{q^{r^2+rs+s^2+st+t^2}}{(q)_r (q)_s (q)_t} z_1^{r+s} z_2^{s+t} = \sum_{a,...
- 21