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awarded Notable Question
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6
awarded Yearling
Jan
6
awarded Yearling
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awarded Nice Question
Nov
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awarded Necromancer
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awarded Benefactor
Nov
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accepted $n$-words over the alphabet $\{0,...,d\}$ without consecutive $0$'s
Nov
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revised $n$-words over the alphabet $\{0,...,d\}$ without consecutive $0$'s
deleted 16 characters in body
Nov
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reviewed Approve suggested edit on Showing that $\sum{x^k}$ does not converge uniformly on $(-1,1)$
Nov
6
reviewed Approve suggested edit on If $A^2 = I$, then $A$ is diagonalizable, and is $I$ if $1$ is its only eigenvalue
Nov
6
comment $n$-words over the alphabet $\{0,...,d\}$ without consecutive $0$'s
But $\sum_{\geq 1} f(n)x^n=\frac{3x+2x^2}{1-2x-2x^2}$, not what you claimed. If you expand your generating function, you see it only agrees with the first two terms.
Nov
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revised $n$-words over the alphabet $\{0,...,d\}$ without consecutive $0$'s
added 13 characters in body
Nov
5
asked $n$-words over the alphabet $\{0,...,d\}$ without consecutive $0$'s
Sep
13
comment Identity involving Stirling numbers of second kind $S(n,k)$ and $k$-compositions of $n$
Thanks for the solution. I didn't even think about the generating function of $S(n,k)$. Once you know that, the problem falls into place.
Sep
13
accepted Identity involving Stirling numbers of second kind $S(n,k)$ and $k$-compositions of $n$
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revised Identity involving Stirling numbers of second kind $S(n,k)$ and $k$-compositions of $n$
deleted 1 character in body
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12
asked Identity involving Stirling numbers of second kind $S(n,k)$ and $k$-compositions of $n$
Jul
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comment How do I approach this combinatorics problem about firing order in an engine?
I believe the answer is $2\cdot n!^2$, as one must first decide whether to start with an even or an odd number.
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awarded Popular Question
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awarded Explainer
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