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 A201732 a(n) = [x^n/n!] (1/x) * log( (n+1 - n*exp(x)) / (n+2 - (n+1)*exp(x)) ). 1

%I #5 Mar 30 2012 18:37:33

%S 1,2,18,386,15150,946082,86148762,10776331778,1773210244230,

%T 371367615732002,96462262816769586,30433572793375652738,

%U 11463680237091180885150,5081782052880868302982562,2618864991559576227420716490,1552537179057766207300655437826

%N a(n) = [x^n/n!] (1/x) * log( (n+1 - n*exp(x)) / (n+2 - (n+1)*exp(x)) ).

%C The function log((n+1 - n*exp(x))/(n+2 - (n+1)*exp(x))) equals the (n+1)-th iteration of log(1/(2-exp(x)), the e.g.f. of A000629 (with offset 1), where A000629(n) is the number of necklaces of partitions of n+1 labeled beads.

%F a(n) = A201731(n+1) / (n+1).

%o (PARI) {a(n)=n!*polcoeff((1/x)*log((n+1 - n*exp(x+O(x^(n+2))))/(n+2 - (n+1)*exp(x+O(x^(n+2))))),n)}

%Y Cf. A201731, A000629.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Dec 04 2011

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Last modified August 14 15:40 EDT 2024. Contains 375165 sequences. (Running on oeis4.)