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A056541 a(n) = 2n*a(n-1) + 1 with a(0)=0.

%I A056541 #32 Jul 02 2025 16:02:00
%S A056541 0,1,5,31,249,2491,29893,418503,6696049,120528883,2410577661,
%T A056541 53032708543,1272785005033,33092410130859,926587483664053,
%U A056541 27797624509921591,889523984317490913,30243815466794691043
%N A056541 a(n) = 2n*a(n-1) + 1 with a(0)=0.
%C A056541 if s(n) is a sequence defined as s(0)=x, s(n) = 2n*s(n-1)+k, n>0, then s(n) = 2^n*n!*x + a(n)*k. - _Gary Detlefs_, Feb 20 2010
%H A056541 Harvey P. Dale, <a href="/A056541/b056541.txt">Table of n, a(n) for n = 0..403</a>
%F A056541 a(n) = floor[(sqrt(e)-1)*2^n*n! ] = A010844(n)-A000165(n).
%F A056541 a(n) = Sum[P(n, k) * 2^k {k=0 to n-1}] - _Ross La Haye_, Sep 15 2004
%F A056541 Conjecture: a(n) +(-2*n-1)*a(n-1) +2*(n-1)*a(n-2)=0. - _R. J. Mathar_, May 29 2013
%F A056541 E.g.f.: (exp(x)-1)/(1-2*x) = -12*x/(Q(0)+6*x-3*x^2)/(1-2*x), where Q(k) = 2*(4*k+1)*(32*k^2+16*k+x^2-6) - x^4*(4*k-1)*(4*k+7)/Q(k+1) ; (continued fraction). - _Sergei N. Gladkovskii_, Nov 18 2013
%e A056541 a(3) = 2*3*a(2)+1 = 6*5+1 = 31.
%t A056541 nxt[{n_,a_}]:={n+1,2a(n+1)+1}; NestList[nxt,{0,0},20][[All,2]] (* or *) With[{nn=20},CoefficientList[Series[(Exp[x]-1)/(1-2x),{x,0,nn}],x] Range[ 0,nn]!] (* _Harvey P. Dale_, Aug 08 2021 *)
%Y A056541 Cf. A002627, A173516.
%K A056541 easy,nonn
%O A056541 0,3
%A A056541 _Henry Bottomley_, Jun 20 2000
%E A056541 More terms from _James Sellers_, Jul 04 2000