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23h

comment 
Combinatorial Proof of Real Analysis Identity
I believe that this identity is a limit of a finite sum that can be proved using the WZ method. Whether that makes it combinatorial is a matter of opinion. 
23h

answered  Identity involving shifted Legendre coefficients 
23h

revised 
Identity involving shifted Legendre coefficients
added 2 characters in body 
2d

comment 
Solving a recurrence (with the form of a convolution) involving binomial coefficients
The identity is just a special case of Vandermonde's theorem: \begin{align*}\sum_{i=1}^{j+1} (1)^{i1}\binom{n+i1}{i1}\binom{n+j+1}{j+1i} & = \sum_{i=1}^{j+1} \binom{n1}{i1}\binom{n+j+1}{j+1i}\\ & = \binom{j}{j}=1. \end{align*} 
Jul 17 
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Combinatorial interpretation of composition of power series?
There is a very nice combinatorial interpretation of identities $g(f(t))=t$ in terms of trees, which is related to operas (which I don't know anything about), due by S. Parker (but not published) and rediscovered by J.L. Loday. (See also R. Bacher and R. Bacher and G. Schaeffer.) However, I couldn't get it to work for this problem. 
Jul 17 
comment 
Given A set $U$ and a set $\mathcal O$ of subsets of $U$, how many subsets of $\mathcal O$ have union $U$?
I don't know anything about the general problem. 
Jul 17 
comment 
Combinatorial interpretation of composition of power series?
This was observed by Michael Somos in 2004, as noted in the OEIS entry. As also noted in the OEIS entry, $f(t) =  tc(t)^3$, where $c(t)$ is the generating function for Catalan numbers. More generally, the compositional inverse of $xc_r(x^a)^b$ is $xc_{abr+1}(x^a)^b$, where $c_r(x)$ is the generalized Catalan number generating function satisfying $c_r(x) = 1+xc_r(x)^r$; the OP's formula is the case $a=1$, $b=3$, $r=2$. 
Jul 17 
answered  Given A set $U$ and a set $\mathcal O$ of subsets of $U$, how many subsets of $\mathcal O$ have union $U$? 
Jul 9 
awarded  Necromancer 
Jul 9 
answered  Combinatorial Morse functions and random permutations 
Jul 8 
comment 
Combinatorial Morse functions and random permutations
I believe that the upper bound should be $L\leq (405581/10!)^{1/10} = 0.80321\cdots$, which may be compared to the actual value $L=0.7693323708\cdots$. 
Jun 3 
comment 
How is this combinatorial structure called?
As far as I'm concerned you can. 
Jun 3 
comment 
How is this combinatorial structure called?
I don't understand what you are asking. 
Jun 2 
answered  How is this combinatorial structure called? 
Jun 2 
revised 
How is this combinatorial structure called?
added 4 characters in body 
Jun 1 
comment 
Counting chains of inclusions
I added the word "exponential". 
Jun 1 
revised 
Counting chains of inclusions
added 12 characters in body 
May 30 
revised 
Counting chains of inclusions
added 1 character in body 
May 30 
revised 
Counting chains of inclusions
added 24 characters in body 
May 30 
answered  Counting chains of inclusions 