Off topic: Can someone lead me to (link) an example of a geometrical simplicial complex? I don't understand what it means to intersect two simplices (and hence how to check such an intersection is a face).
does this simply mean that the intersections can only occur along vertices, edges, and faces?
every summand is $23!/15!$, so we must multiply that by the number of terms, which is the number of ways to choose $f(2)$ and $f(4)$ according to the conditions, which is $4\cdot 11+10\cdot12$. so your answer is correct, Peter.
for f(2) in {9,10,11,12}, there are 12-2=10 choices for f(4) (we must exclude both f(2) and 13-f(2)), otherwise there are 12 choices for f(4) if f(2)>12, so the count should be 23!/15! times 4*10+10*12.
for that, you need to find the highest $n$ for which $21^2-4\cdot7-2^n$ is a square mod $2^5$, for which it suffices to just run down the list $n=5,4,3,\cdots$.
($a^2+21a+7\equiv 2^n$ mod $2^5$ has a solution iff $21^2-4\cdot7-2^n$ is square mod $2^5$)
note that $\Delta$ is a square mod $2^{{\rm blah}\ge3}$ iff it is a square mod $2^3$
oh, it would have to be something like $2^nm$ as $m$ can vary too, drats
the point is to see just how many times 2 can divide into the residue mod 2^5
that will be the power of 2 in the resulting gcd
list out the squares mod 2^5 (these are 0,1,9,17,25,4,16), translate this set additively by 4*7-21^2, then look at the 2-adic valuations of the resulting residues, I think
@robjohn The chat in that question is going to be useful in the future - Did you see the questions I mentioned? Specially the one in mathematica.se? I can use that chat to show Artes why I didn't accept his answer. xD
@Ethan I have something nice to you $$\space \lim_{n\to\infty}\sum_{k=0}^{n} \frac{\cosh (k \pi /n)}{(k+1)^2}$$ I know you like these things ;) - - - (and it's created by me)
