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Jan
22
comment Curious Binomial Coefficient Identity
it's a (yet another) form of Vandermonde's identity
Jan
20
reviewed Close How to understand the regular cardinal?
Jan
20
reviewed Approve mantel theorem bipartite graphs, two triangles share an edge
Jan
19
reviewed Close I need help to evaluate this definite integral.
Jan
19
reviewed Close Solve 10 base logarithms
Jan
19
reviewed Close Forming equations for exponential growth/decay questions
Jan
19
comment Binomial Identity $\sum\binom{2n+1}{2k+1}\binom{m+k}{2n} = \binom{2m}{2n}$
see also math.stackexchange.com/q/1107465 for a bijective proof of an equivalent identity
Jan
19
comment Showing ${n + a - 1 \choose a - 1} = \sum_{k = 0}^{\left\lfloor n/2 \right\rfloor} {a \choose n-2k}{k+a-1 \choose a-1}$
come to think of it, substituting $a\to 2n+1$, $n\to 2m-2n$ (and $k\to m-2n+k$) one can see that these two identities are equivalent
Jan
19
comment Showing ${n + a - 1 \choose a - 1} = \sum_{k = 0}^{\left\lfloor n/2 \right\rfloor} {a \choose n-2k}{k+a-1 \choose a-1}$
Both identities are of the form $\sum_k\binom a{b-2k}\binom{c+k}d=\binom ef$ — and I strongly suspect there is a unified proof.
Jan
19
reviewed Looks OK Let $G$ be a group, and $H$ a subgroup of $G$. Let $a, b \in G$. Prove $Ha=Hb$ iff $ab^{-1} \in H$.
Jan
19
reviewed Leave Closed An endless loop in a program that search for mathematical theorems and proofs − a milestone?
Jan
19
reviewed Close Min, Max, Infimum, Supremum
Jan
19
reviewed Leave Open Given a general 3D Matrix operation … who can I apply “1/2” of the effect of it ?
Jan
19
reviewed Looks OK Find prime factorization of $2^{22} + 1$
Jan
19
reviewed Close cohomology ring of a quotient space
Jan
18
reviewed Close Simplify the radical expression
Jan
18
reviewed Reject Fallacy of denying the hypothesis
Jan
18
reviewed Reject Need function for tunable sigmoid 2D surface
Jan
18
reviewed Close Why integration by part (not partial) is considered everywhere useful?
Jan
18
reviewed Leave Open How prove this limits $\lim_{n\to\infty}\frac{v_{5}(1^1\cdot 2^2\cdot 3^3\cdot 4^4\cdots\cdot n^n)}{n^2}=\frac{1}{8}$