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Jan
16
comment On ${-1 \choose 0}=1$, can I assume that $\frac{(-1)!}{(-1)!}=1$?
OK, I apologise — this question was not completely clear as written — but it's a) not a duplicate of the linked question; b) can be reasonably answered.
Jan
16
comment On ${-1 \choose 0}=1$, can I assume that $\frac{(-1)!}{(-1)!}=1$?
related: How to use $\binom a k = \frac{\alpha(a-1)(a-2)\cdots(a-k+1)}{k(k-1)(k-2)\cdots 1}$ to check that ${-1\choose 0}=1$?
Jan
15
comment Unimodality of q-binomial coefficients
Unimodality of q-binomial coefficients is a difficult theorem, proved more than 20 years after it was conjectured. If you're really interested in a proof — it's easy to google references.
Jan
15
comment Finite summation with binomial coefficients, $\sum (-1)^k\binom{r}{k} \binom{k/2}{q}$
@655321 though there is one example that is explained in detail in many books (e.g. in Enumerative Combinatorics): Lagrange inversion formula
Jan
15
comment Finite summation with binomial coefficients, $\sum (-1)^k\binom{r}{k} \binom{k/2}{q}$
@655321 I'm afraid I don't know a good reference...
Jan
14
comment Embedding of two-dimensional CW complexes which induces a zero homomorphism on second homotopy groups
x-posted to MO: mathoverflow.net/q/193845
Jan
14
comment When $\frac{C(n, k)}{n^{k-1}} > 1$?
Well, for $n=k!+t$ we need to compare $(1+t/k!)\cdot(1-1/(k!+t))(1-2/(k!+t))\ldots(1-(k-1)/(k!+t))$ withs 1. Shouldn't be hard (take logarithm, bound it...) — and at least genesis of $s_k$ is clear.
Jan
13
comment Summation with Binomial Coefficients, $\sum (-1)^k \binom{m_1}{k} \binom{m_2}{k} $
cf. math.stackexchange.com/q/858250
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?