All Questions
25 questions
0
votes
1
answer
129
views
Closed form for $\sum\limits_{k=0}^{n} [\operatorname{wt}(k) = m]$ where $\operatorname{wt}(n)$ is the binary weight of $n$
Let $\operatorname{wt}(n)$ be A000120 (i.e., number of $1$'s in binary expansion of $n$).
Let $a(n,m)$ be the family of integer sequences such that
$$
a(n,m) = \sum\limits_{k=0}^{n} [\operatorname{wt}(...
- 4,938
0
votes
1
answer
169
views
Partial sums of binomial coefficients and related family of polynomials
Let $a(n)$ be A302117. Here
$$
a(n) = 4(n-1)a(n-1) - \frac{1}{3}\prod\limits_{k=0}^{n-1}(2k-3), \\
a(0) = 0.
$$
Let
$$
T(n,k) = \sum\limits_{i=0}^{k} \binom{n}{i}.
$$
Let $P_n(z)$ be the family of ...
- 4,938
6
votes
0
answers
752
views
For all $n\in \mathbb{N}$, How to find $\min\{ m+k\}$ such that $ \binom{m}{k}=n$?
I asked this question on MSE here.
Most numbers in pascal triangle appear only once (excluding the duplicates in the same row of the Pascal's triangle) but certain numbers appear multiple times. ...
- 541
1
vote
0
answers
73
views
Alternating sum of integer coefficients of the triangles related to Eulerian numbers and binomial transforms
Let $W(n, k, m)$ be an integer coefficients defined for $n > 0, 1 \leqslant k \leqslant n, m > 0$ with $W(n,k,m)=0$ for $n \leqslant 0$ or $k \leqslant 0$ such that
$$
W(n, k, m) = (k+m-1)W(n-1,...
- 4,938
7
votes
1
answer
527
views
Suitable closed form for the A079501
Let $a(n)$ be A079501 (i.e., number of compositions of the integer $n$ with strictly smallest part in the first position).
The sequence begins with
$$
1, 1, 2, 2, 4, 5, 8, 12, 19, 28, 45, 70, 110, ...
- 4,938
13
votes
1
answer
468
views
Four new series for $\pi$ and related identities involving harmonic numbers
Recently, I discovered the following four new (conjectural) series for $\pi$:
\begin{align}\sum_{k=1}^\infty\frac{(5k^2-4k+1)8^k\binom{3k}k}{k(3k-1)(3k-2)\binom{2k}k\binom{4k}{2k}}&=\frac{3\pi}2,\...
- 15.6k
1
vote
0
answers
95
views
On level-$12$ of the McKay-Thompson series of the Monster and the Domb numbers
(This continues from level 10.) Given some moonshine functions $j_{N}$. There are nice descending and consistent relations for levels $6m$ with $m= 2,3,5,$
$$j_{12A} = \left(\sqrt{j_{12H}} + \frac{\...
- 12.6k
1
vote
0
answers
116
views
On level $6$ of the McKay–Thompson series of the Monster and Apéry numbers, et al
After the McKay–Thompson series of levels $1$, $2$, $3$, $4$ of the Monster were mentioned in this MO post, level $6$ has very interesting relations as well. (Level 10 is in this post.)
I. Level-6 ...
- 12.6k
2
votes
0
answers
70
views
Integer coefficients such that $T(n,k)=R(n,k)-R(n,k-1)$
Let $a(n)$ be A000085, i.e., the number of self-inverse permutations on $n$ letters, also known as involutions; number of standard Young tableaux with $n$ cells. Here
$$a(n) = a(n-1) + (n-1)a(n-2), a(...
- 4,938
10
votes
1
answer
435
views
Series for $\frac{\log m}{\pi}$ with summands involving harmonic numbers
The classical rational Ramanujan-type series for $1/\pi$ have the following four forms:
\begin{align}\sum_{k=0}^\infty(ak+b)\frac{\binom{2k}k^3}{m^k}&=\frac{c}{\pi},\label{1}\tag{1}
\\\sum_{k=0}^\...
- 15.6k
4
votes
0
answers
279
views
What is the exact value of the series $\sum_{k=0}^\infty \binom{2k}k^4/256^k$?
By Stirling's formula $n!\sim\sqrt{2\pi n}(n/e)^n$, we have
$$\binom{2k}k\sim\frac{4^k}{\sqrt{k\pi}}$$
and hence
$$\frac{\binom{2k}k^4}{256^k}\sim\frac1{k^2\pi^2}.\tag{1}$$
So the series
$$\sum_{k=0}^\...
- 15.6k
2
votes
0
answers
219
views
Question on globally convergent formulas for the Riemann zeta function $\zeta(s)$
Consider the following two formulas for $\zeta(s)$
$$\zeta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{1-2^{1-s}}\sum\limits_{n=0}^K \frac{1}{2^{n+1}}\sum\limits_{k=0}^n \binom{n}{k} \frac{(-1)^...
- 1,126
2
votes
0
answers
215
views
Two conjectures about generalised A329369
Let $m \geqslant 2$ be a fixed integer.
Let
$$\operatorname{wt}(n,m)=\operatorname{wt}\left(\left\lfloor\frac{n}{m}\right\rfloor,m\right)+n\bmod m, \operatorname{wt}(0,m)=0$$
Then we have an integer ...
- 4,938
2
votes
1
answer
113
views
Modulo $2$ binomial transform of A243499 applied $k$ times
Let $m \geqslant 1$ be a fixed integer.
Let $f(n)$ be A007814, exponent of highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$.
...
- 4,938
0
votes
1
answer
149
views
Modulo $2$ binomial transform of A124758
Let $f(n)$ be A153733, remove all trailing ones in binary representation of $n$. Here
\begin{align}
f(2n)& = 2n\\
f(2n+1)& = f(n)\\
\end{align}
Then we have an integer sequence given by
\begin{...
- 4,938
1
vote
0
answers
57
views
Inverse modulo $2$ binomial transform of generalised A284005
Let $m \geqslant 1$ be a fixed integer.
Let $\operatorname{wt}(n)$ be A000120, $1$'s-counting sequence: number of $1$'s in binary expansion of $n$ (or the binary weight of $n$).
Let $f(n)$ be A007814, ...
- 4,938
1
vote
0
answers
156
views
Open tours by a biased rook (proof verification)
Related questions:
Number of open tours by a biased rook on a specific $f(n)\times 1$ board which end on a $k$-th cell from the right
Sum with products turned into subsequences
Combinatorial ...
- 4,938
2
votes
2
answers
180
views
Modulo $2$ binomial transform of $m^n$
Let $m \in \mathbb{R}$.
Let $f(n)$ be A007814, exponent of highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$.
Let $g(n)$ be ...
- 4,938
3
votes
0
answers
151
views
Combinatorial interpretation of inverse modulo $2$ binomial transform of A284005
My question is related to the following:
Sum with products turned into subsequences
We have an identity
$$a(n, -1) = \sum\limits_{j=0}^{2^{\operatorname{wt}(n)}-1}(-1)^{\operatorname{wt}(n)-\...
- 4,938
2
votes
1
answer
290
views
Evaluations of three series involving binomial coefficients
Question. How to prove the following three identities?
\begin{align}\sum_{k=1}^\infty\frac1{k(-2)^k\binom{2k}k}\left(\frac1{k+1}+\ldots+\frac1{2k}\right)=\frac{\log^22}3-\frac{\pi^2}{36},\tag{1}
\end{...
- 15.6k
1
vote
0
answers
154
views
An explicit solution to the congruence $x^2\equiv 14(\frac 3p)-(\frac p3)-12\pmod {p}$?
Let $p>3$ be a prime, and let $(\frac{\cdot}p)$ be the Legendre symbol. Then
$$14\left(\frac 3p\right)-\left(\frac p3\right)-12=\begin{cases}1&\text{if}\ p\equiv1\pmod{12},
\\-25&\text{if}\ ...
- 15.6k
2
votes
1
answer
194
views
Does this series, related to the Hasse/Ser series for $\zeta(s)$, converge for all $s \in \mathbb{C}$?
I have asked this question at math stack exchange, however it did not get any traction. Still curious about the answer though.
Numerical evidence suggests that:
$$\lim_{N \to +\infty} \sum_{n=1}^N\...
- 4,169
7
votes
1
answer
439
views
Two conjectural series for $\pi$ involving the central trinomial coefficients
For each $n=0,1,2,\ldots$, the central trinomial coefficient $T_n$ is defined as the coefficient of $x^n$ in the expansion of $(x^2+x+1)^n$. It is easy to see that $T_n=\sum_{k=0}^{\lfloor n/2\rfloor}\...
- 15.6k
37
votes
3
answers
2k
views
How to prove the identity $L(2,(\frac{\cdot}3))=\frac2{15}\sum\limits_{k=1}^\infty\frac{48^k}{k(2k-1)\binom{4k}{2k}\binom{2k}k}$?
For the Dirichlet character $\chi(a)=(\frac a3)$ (which is the Legendre symbol), we have
$$L(2,\chi)=\sum_{n=1}^\infty\frac{(\frac n3)}{n^2}=0.781302412896486296867187429624\ldots.$$
Note that this ...
- 15.6k
18
votes
5
answers
3k
views
Bernoulli sum meets golden number
Let $B_n$ denote the Bernoulli numbers and let $\phi=\frac{1+\sqrt{5}}2$ be the golden ratio.
I encountered the following infinite sum and would like to ask:
Question. Is this true? If so, any ...
- 43.2k