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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...)
Jan
3
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
Jan
3
comment More on primes $p=u^2+27v^2$ and roots of unity
(and math.stackexchange.com/q/31485 is, of course, related)
Jan
3
comment More on primes $p=u^2+27v^2$ and roots of unity
cf. statement in the upd. of math.stackexchange.com/q/102736
Jan
3
comment Manifold with $\pi_1(M)=F_n$
when you say 'manifold' you mean compact w/o boundary, I guess?
Jan
3
comment What is the least number $n$, such that $n^{2015}+2015$ is prime?
related: math.stackexchange.com/q/663884
Jan
3
comment Show that $\mathbb{R} P^3$ is not homotopy equivalent to $\mathbb{R} P^2 \vee S^3$.
...or compute $\pi_2$
Jan
3
comment How do I evaluate $\sum_{r=1}^{n} [r(r+1)(r+2)(r+3)] $?
Compute answers for small $n$. Then try to guess the general answer. Then try to prove it.
Jan
3
comment A counter example in obstruction theory
@Qiaochu This was x-posted to MO and answered there ($X=\mathbb RP^3$, $Y=\mathbb RP^2$)