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Jan
17
comment Showing ${n + a - 1 \choose a - 1} = \sum_{k = 0}^{\left\lfloor n/2 \right\rfloor} {a \choose n-2k}{k+a-1 \choose a-1}$
cf. math.stackexchange.com/q/601940 etc
Jan
17
comment Elementary proof of the fact that any orientable 3-manifold is parallelizable
«The second surprise in this story» — that's, IMHO, the most interesting part...
Jan
17
comment Closed-form solution for $f(n) = \sum_{k>0}\binom{n}{2k}x^{k}$ without $\sqrt{x}$
@1234 $\frac{P(x)+P(-x)}2$ extracts even terms from any polynomial $P$, you know...
Jan
17
comment Closed-form solution for $f(n) = \sum_{k>0}\binom{n}{2k}x^{k}$ without $\sqrt{x}$
Well, do you know the answer for just $\sum \binom nlx^l$?
Jan
17
comment Elementary proof of the fact that any orientable 3-manifold is parallelizable
Do you know the standard proof («$w_1=w_2=0$ implies parallelizable by elementary obstruction theory» + «$w_1=0$ implies $w_2=0$ by Wu's formulas»)? It's not that hard — and at least the first part is, in a sense, the most straightforward approach possible (but I indeed don't know any intuitive explanation of the second part).
Jan
16
comment If $a,b,c$ are real numbers all less than or equal to $1$ such that $a+b+c=0$ , then is it true that $(1-a)(1-b)(1-c) \le 1$?
«which is true» — why, actually? ($ab$ can be negative)
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)