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seen Dec 18 '12 at 3:32

Aug
23
awarded  Nice Question
Dec
17
accepted Relationship between ordered and binary trees
Dec
16
comment Relationship between ordered and binary trees
@Isomorphism: I am looking for a different bijection. The post you linked is about a bijection for full binary trees, whereas I require a bijection for a particular number of left and right children.
Dec
15
asked Relationship between ordered and binary trees
Dec
13
accepted permutation with cycles
Dec
13
comment permutation with cycles
Thank you very much!
Dec
13
comment Simon Newcomb's problem
For 1. I somehow have to use that $A_{\emptyset}(t)=1$ and $\frac{A_S(t)}{(1-t)^{d+1}}=\sum\limits_{n=0}^{\infty}t^n\prod\limits_i\binom{n+‌​d_i-1}{d_i}$
Dec
13
asked Simon Newcomb's problem
Dec
13
comment permutation with cycles
Thank you very much for your answer. Unfortunately I cannot find the sketch of the proof you were talking about.
Dec
11
asked permutation with cycles
Dec
7
awarded  Commentator
Dec
7
comment Generating function for binomial coefficients $\binom{2n+k}{n}$ with fixed $k$
I am also looking for a combinatorical proof of this identity.
Nov
27
awarded  Scholar
Nov
27
accepted Combinatorial Interpretation
Nov
27
accepted Generating function for binomial coefficients $\binom{2n+k}{n}$ with fixed $k$
Nov
26
comment Combinatorial Interpretation
Thank you, Dennis. However, I do not have much experience with combinatorics so I am stuck with coming up with ideas for the other two. I know that 1.) it is choosing $s$ elements out of a $(n-m)r$-set and 2.) choosing $r$ elements from a $n-m$-set and then choosing $s$ elements from those. Still I dont see how this can be intersections of sets with bad properties.
Nov
26
comment Generating function for binomial coefficients $\binom{2n+k}{n}$ with fixed $k$
Thank you, but do you have a page number for me, please.
Nov
26
comment Generating function for binomial coefficients $\binom{2n+k}{n}$ with fixed $k$
$$\sum_{n=0}^k \binom{k}{n} \left( \frac{1}{2t} \right)^k (- \sqrt{1-4t})^{k-n-1} (2t)^{-k+n} $$
Nov
25
comment Combinatorial Interpretation
So $f(n)$ is the number of graphs on n vertices without isolated vertices because $2^{\binom{k}{2}}$ is the number of elements in the intersection of all sets of labeled graphs on $n$ vertices with an isolated vertex?!
Nov
24
comment Generating function for binomial coefficients $\binom{2n+k}{n}$ with fixed $k$
I expanded the left hand side to $\sum_{n=0}^k\binom{k}{n}\left(\frac{1}{2t}\right)^k\frac{-\sqrt{1-4t})^{k-n-1}}‌​{2t}^{k-n}}$ but that's it. I do not know how to proceed.