Grigory M
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 Jan9 comment A three variable binomial coefficient identity I've also asked a (different but) related question @ MO Jan9 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 $EM(n)\to EM(n+1)$. Jan8 comment Kernel and image of a homomorphism $SL(2,5)\to S_5$ Jan7 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? Jan6 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 Jan6 comment K-theory formulation of the index theorem ...and equivalence of these two forms follows from Hirzebruch-Riemann-Roch theorem. Jan6 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. Jan5 comment Binomial Sum Related to Fibonacci: $\sum\binom{n-i}j\binom{n-j}i=F_{2n+1}$ Jan5 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 $m=n$... Jan5 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. Jan5 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). Jan4 comment A three variable binomial coefficient identity also triple product looks superficially similar to 3-variable form of Dixon / Strehl Jan3 comment More on primes $p=u^2+27v^2$ and roots of unity Explicit formulas would be very nice indeed... Jan3 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 Jan3 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 Jan3 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...) Jan3 comment More on primes $p=u^2+27v^2$ and roots of unity and your $k$ is always an element of order $(p-1)/3$ in $\mathbb Z/p^\times$, I guess Jan3 comment More on primes $p=u^2+27v^2$ and roots of unity (and math.stackexchange.com/q/31485 is, of course, related) Jan3 comment More on primes $p=u^2+27v^2$ and roots of unity cf. statement in the upd. of math.stackexchange.com/q/102736 Jan3 comment Manifold with $\pi_1(M)=F_n$ when you say 'manifold' you mean compact w/o boundary, I guess?