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Jan
13
comment Solve easy sums with Binomial Coefficient
possible duplicate of Understanding a combinatorial relation.
Jan
13
comment Summation with Binomial Coefficients, $\sum (-1)^k \binom{m_1}{k} \binom{m_2}{k} $
related: $\sum(-1)^k\binom mk^2$
Jan
12
comment Groups with no nontrivial topology
(Re: «I am quite interested in how the problem changes if G is infinite or finite.») That's easy to answer: changes from completely trivial (finite) to very hard problem that stayed open for almost 40 years (infinite).
Jan
12
comment Groups with no nontrivial topology
@ MO: mathoverflow.net/questions/165783/…
Jan
10
comment Are Exponential and Trigonometric Functions the Only Non-Trivial Solutions to $F'(x)=F(x+a)$?
related: math.stackexchange.com/q/199691
Jan
9
comment Finite summation with binomial coefficients, $\sum (-1)^k\binom{r}{k} \binom{k/2}{q}$
off the top of my head, something like math.stackexchange.com/a/609202 should work (but finding a bijective proof would be, perhaps, more challenging)
Jan
9
comment Finite summation with binomial coefficients, $\sum (-1)^k\binom{r}{k} \binom{k/2}{q}$
@Marko tomorrow (or later) — maybe; if you have time now — please just go ahead
Jan
9
comment A three variable binomial coefficient identity
Thank you! I'm awarding the bounty now — and will try to understand the proof later.
Jan
9
comment A three variable binomial coefficient identity
I've also asked a (different but) related question @ MO
Jan
9
comment Is this morphism of spectra zero in the stable homotopy category?
No, that generalization is not true: take any non-trivial stable cohomological operation — say, Bockstein $EM(n)\to EM(n+1)$.
Jan
8
comment Kernel and image of a homomorphism $SL(2,5)\to S_5$
related: Elementary isomorphism between $PSL(2,5)$ and $A_5$ etc
Jan
7
comment A three variable binomial coefficient identity
I now suspect that both sides count 00-avoiding $3n$-periodic binary sequences with exactly $n$ zeroes — maybe someone can prove it?
Jan
6
comment K-theory formulation of the index theorem
as for the last paragraph, it's really too broad — but perhaps math.stackexchange.com/q/295050 is related
Jan
6
comment K-theory formulation of the index theorem
...and equivalence of these two forms follows from Hirzebruch-Riemann-Roch theorem.
Jan
6
comment $S^1 \vee S^1 \vee S^2$ and $S^1 \times S^1$ have same homology but are not homotopy equivalent
Related: Two CW complexes with isomorphic homotopy groups and homology, yet not homotopy equivalent
Jan
6
comment Do you decline a multiplier in reading a mathematical formula in Russian?
Dear user204305, I'm glad if my comments help — but it wouldn't be appropriate for me to post an answer to a question I consider offtopic.
Jan
5
comment Binomial Sum Related to Fibonacci: $\sum\binom{n-i}j\binom{n-j}i=F_{2n+1}$
cf. $\sum_{i+j+k=n}\binom{n-k}i\binom{n-i}j\binom{n-j}k$
Jan
5
comment Binomial Identity $\sum\binom{2n+1}{2k+1}\binom{m+k}{2n} = \binom{2m}{2n}$
I wonder if there is a short proof using the fact that both sides are polynomials in $m$ of deg $2n$ that are both zero in $m=0,1,\ldots,n-1$ and are equal for $m=n$...
Jan
5
comment Do you decline a multiplier in reading a mathematical formula in Russian?
In this case too a purist would insist that only one variant is correct («двум эн») but in practice both variants are frequently used.
Jan
5
comment Do you decline a multiplier in reading a mathematical formula in Russian?
В принципе, пурист сказал бы «цэ равно сумме а-пятого и а-шестого». Но часто говорят «цэ равно а-пять плюс а-шесть» (что, разумеется, является некоторым жаргонизмом).