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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 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).
Jan
4
comment A three variable binomial coefficient identity
also triple product looks superficially similar to 3-variable form of Dixon / Strehl
Jan
3
comment More on primes $p=u^2+27v^2$ and roots of unity
Explicit formulas would be very nice indeed...
Jan
3
comment Ramanujan-type trigonometric identities with cube roots, generalizing $\sqrt[3]{\cos(2\pi/7)}+\sqrt[3]{\cos(4\pi/7)}+\sqrt[3]{\cos(8\pi/7)}$
Related: upd. in math.stackexchange.com/q/102736 and math.stackexchange.com/a/31600
Jan
3
comment Ramanujan-type trigonometric identities with cube roots, generalizing $\sqrt[3]{\cos(2\pi/7)}+\sqrt[3]{\cos(4\pi/7)}+\sqrt[3]{\cos(8\pi/7)}$
I'll try to expand this answer later
Jan
3
comment More on primes $p=u^2+27v^2$ and roots of unity
(Looks like we already have all pieces of the puzzle and they start to fit together... — but I haven't yet solved it...)