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Well it's me, Marko Riedel. ;-)

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10h
revised How many “good” graphs of size $n$ are there?
proper index
14h
comment How many “good” graphs of size $n$ are there?
The history of this method including of course the work by Flajolet and Sedgewick and the notation are documented at this Wikipedia entry.
14h
revised How many “good” graphs of size $n$ are there?
treat the case of no fixed points
14h
revised How many connected graphs over V vertices and E edges?
linkage
1d
revised How many “good” graphs of size $n$ are there?
wikipedia
1d
answered How many “good” graphs of size $n$ are there?
1d
comment How many connected graphs over V vertices and E edges?
That perhaps was a less than ideal choice of variables on my part. Do try to read the derivation and the Maple code and it should become clear that the $q_{n,k}$ on the left is the count of labeled connected graphs with $k$ edges and on $n$ nodes while the one on the right is the length of the partition $\lambda.$ Observe that the third method is better than the second one.
2d
answered Prove the following relation:
2d
comment Counting the number of unicyclic graphs
Labelled unicyclic graphs are counted at this MSE link.
Dec
16
answered How many connected graphs over V vertices and E edges?
Dec
15
revised Binomial transform
addendum
Dec
15
answered Binomial transform
Dec
15
answered Harmonic number identity
Dec
14
comment Compute $\int_0^\infty \frac{dx}{x^5+1}$ using a contour in the upper half complex plane that encloses one of the roots of $z^5+1=0$
The following MSE link may prove useful reading.
Dec
13
revised How to prove combinatorial identity $\sum_{k=0}^s{s\choose k}{m\choose k}{k\choose m-s}={2s\choose s}{s\choose m-s}$?
add linkage
Dec
13
revised How prove binomial cofficients $\sum_{k=0}^{[\frac{n}{3}]}(-1)^k\binom{n+1}{k}\binom{2n-3k}{n}=\sum_{k=[\frac{n}{2}]}^n\binom{n+1}{k}\binom{k}{n-k}$
missing differential
Dec
13
answered How prove binomial cofficients $\sum_{k=0}^{[\frac{n}{3}]}(-1)^k\binom{n+1}{k}\binom{2n-3k}{n}=\sum_{k=[\frac{n}{2}]}^n\binom{n+1}{k}\binom{k}{n-k}$
Dec
13
comment How to prove combinatorial identity $\sum_{k=0}^s{s\choose k}{m\choose k}{k\choose m-s}={2s\choose s}{s\choose m-s}$?
It is debatable whether one should write a new introduction for similar problems, I do copy from my previous posts sometimes when I write new ones, with the source already being present in my editor.
Dec
13
revised How prove this identity$\sum_{k=0}^{n}\binom{2k}{k}\binom{n+k}{2k}(s-t)^{n-k}t^k=\sum_{k=0}^{n}\binom{n}{k}^2s^{n-k}t^k$
linkage
Dec
13
comment How prove this identity$\sum_{k=0}^{n}\binom{2k}{k}\binom{n+k}{2k}(s-t)^{n-k}t^k=\sum_{k=0}^{n}\binom{n}{k}^2s^{n-k}t^k$
When you ask Maple it just collapses the binomial to where you started from. Owing to the dependence on $s$ and $t$ you cannot conclude that it is equal to ${n\choose k}.$ There are many choices for the red term that would give the same value of the sum.