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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
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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)^{i-1}\binom{n+i-1}{i-1}\binom{n+j+1}{j+1-i} & = \sum_{i=1}^{j+1} \binom{-n-1}{i-1}\binom{n+j+1}{j+1-i}\\ & = \binom{j}{j}=1. \end{align*}
Jul
17
comment 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 $xc_{ab-r+1}(-x^a)^b$, 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?
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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
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May
30
revised Counting chains of inclusions
added 24 characters in body
May
30
answered Counting chains of inclusions