All Questions
15 questions with no upvoted or accepted answers
7
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Some conjectural congruences involving Domb numbers
The Domb numbers are given by
$$D_n=\sum_{k=0}^n\binom{n}{k}^2\binom{2k}k\binom{2(n-k)}{n-k}\ \ \ (n=0,1,2,\ldots).$$
Such numbers have combinatorial interpretation, see, e.g., http://oeis.org/A002895....
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7
votes
0
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264
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Is every integer $n>1$ the sum of two squares and two central binomial coefficients?
Those integers $\binom{2n}n\ (n=0,1,2,\ldots)$ are called central binomial coefficients. By Stirling's formula,
$$\binom{2n}n\sim \frac{4^n}{\sqrt{n\pi}}\ \ \ \ (n\to+\infty).$$
Of course, the ...
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7
votes
0
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388
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Polynomials with presumably positive coefficients
The $q$-Pochhammer symbol
$(q) _ 0:=1$ and $(q) _ n:=\prod _ {j=1}^n (1-q^j)$ for $n > 0$ is clearly a polynomial in $q$ which has both positive and negative coefficients when $n>0$.
The $q$-...
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6
votes
0
answers
752
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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. ...
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6
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0
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217
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Two conjectural congruences for Franel numbers
Recall that the Franel numbers are given by
$$f_n:=\sum_{k=0}^n \binom{n}{k}^3\ \ \ (n=0,1,\ldots).$$
Question. How to prove my following conjecture?
Conjecture. For each odd prime $p$, we have
$$\...
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3
votes
0
answers
144
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Flat polynomials with factors of big height
Let $p(x)$ be a polynomial of degree $n$ with all coefficients in $\{-1,0,1\}$ (such polynomials are sometimes called flat). I am wondering how big the coefficients of a factor of $p$ can be. Call ...
- 13.4k
3
votes
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151
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Combinatorial interpretation of inverse modulo $2$ binomial transform of A284005
My question is related to the following:
Sum with products turned into subsequences
We have an identity
$$a(n, -1) = \sum\limits_{j=0}^{2^{\operatorname{wt}(n)}-1}(-1)^{\operatorname{wt}(n)-\...
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2
votes
0
answers
70
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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
$$a(n) = a(n-1) + (n-1)a(n-2), a(...
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2
votes
0
answers
215
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Two conjectures about generalised A329369
Let $m \geqslant 2$ be a fixed integer.
Let
$$\operatorname{wt}(n,m)=\operatorname{wt}\left(\left\lfloor\frac{n}{m}\right\rfloor,m\right)+n\bmod m, \operatorname{wt}(0,m)=0$$
Then we have an integer ...
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1
vote
0
answers
73
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Alternating sum of integer coefficients of the triangles related to Eulerian numbers and binomial transforms
Let $W(n, k, m)$ be an integer coefficients defined for $n > 0, 1 \leqslant k \leqslant n, m > 0$ with $W(n,k,m)=0$ for $n \leqslant 0$ or $k \leqslant 0$ such that
$$
W(n, k, m) = (k+m-1)W(n-1,...
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1
vote
0
answers
57
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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, ...
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1
vote
0
answers
156
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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 ...
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0
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0
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86
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How to prove the following equation (involving multiple binomial coefficients sum)?
I encountered the equation below, encountered a problem that has been bothering me for a long time
Does anyone have an idea how to prove it? I would be extremely grateful to you if you come up with an ...
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0
votes
0
answers
128
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Number of primes skipped by binomial coefficients?
Take $$B(l,n)=\binom{n+l}{n}$$ and $\mathcal P(t)=\{p\mbox{ prime}:p|t\}$.
What is the cardinality of $\mathcal P(B(l,n))$?
What is minimum cardinality of $L\subseteq\{1,\dots,n\}$ such that $$\...
- 13.9k
0
votes
0
answers
369
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A question on the Faulhaber's formula
Proposition 1.1
For every integers $m,n\geq 0$ the following identity holds
\begin{equation}
n^{2m+1}=\sum_{k=1}^{n}\sum_{j=0}^m A_{m,j}k\strut^j(n-k)\strut^j=\sum_{k=0}^{m}(-1)^{m-k}U_m(n, k)\cdot n\...
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