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 Apr 19 awarded Popular Question Apr 7 awarded Nice Answer Apr 7 awarded Nice Question Mar 26 awarded Nice Question Mar 25 awarded Popular Question Jan 28 awarded Notable Question Jan 6 awarded Yearling Nov 16 awarded Nice Question Nov 11 awarded Necromancer Nov 9 awarded Benefactor Nov 8 accepted $n$-words over the alphabet $\{0,…,d\}$ without consecutive $0$'s Nov 7 revised $n$-words over the alphabet $\{0,…,d\}$ without consecutive $0$'s deleted 16 characters in body Nov 6 reviewed Approve Showing that $\sum{x^k}$ does not converge uniformly on $(-1,1)$ Nov 6 reviewed Approve 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 6 revised $n$-words over the alphabet $\{0,…,d\}$ without consecutive $0$'s added 13 characters in body Nov 5 awarded Notable Question 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 $n$