A126674 - OEIS

%I #41 Sep 08 2022 08:45:29

%S 0,1,4,20,128,1024,9984,115968,1572864,24477696,430571520,8452177920,

%T 183175741440,4343275192320,111817607086080,3105593229312000,

%U 92539365359616000,2944365169213440000,99619235621240832000,3571109329517936640000,135199252993504444416000,5390266968989421797376000

%N a(n) = n!*Sum_{j=0..n-1} 2^j/(j+1).

%C _R. J. Mathar_'s recurrence is correct.  a(n) has a new sum term in addition to what a(n-1) has, giving a(n) = n*a(n-1) + 2^(n-1)*(n-1)!.  (Cf. A000165 = 2^n*n!.)  The same for a(n-1) from a(n-2), and a factor, is 2*(n-1)*(a(n-1) - (n-1)*a(n-2)) = 2^(n-1)*(n-1)! too.  Substitute it leaving a(n) in terms of a(n-1) and a(n-2).  The recurrence shows the o.g.f. satisfies the differential equation (2*x^2-x+1)*g + 3*x^2*(2*x-1)*g' + 2*x^4*g'' - x = 0. - _Kevin Ryde_, Jul 11 2019

%H N. J. A. Sloane, <a href="/A126671/a126671.txt">Notes on Carlo Wood's Polynomials</a>

%F From _Vladeta Jovovic_, Feb 13 2007: (Start)

%F a(n) = 2^(n-1)*A003149(n-1).

%F O.g.f.: x*(Sum_{k>=0} k!*(2*x)^k)^2.

%F E.g.f.: log(1-2*x)/(x-1)/2. (End)

%F E.g.f.: E(x) = 1/2*log(1 - 2*x)/(x - 1) = x*(1 - x*G(0))/(x-1)/(2*x-1); G(k) = 1 + 2*x*(2*k+1)/(2*k + 3 - 2*x*(k+1)*(2*k+3)/(2*x*(k+1) + (k+2)/G(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, Dec 13 2011

%F G.f.: x*hypergeom([1,1],[],2*x)^2. - _Mark van Hoeij_, May 16 2013

%F Conjecture: a(n) + (-3*n+2)*a(n-1) + 2*(n-1)^2*a(n-2) = 0. - _R. J. Mathar_, May 23 2014

%F G.f.: x*(1/(1 - 2*x/(1 - 2*x/(1 - 4*x/(1 - 4*x/(1 - 6*x/(1 - 6*x/(1 - ...))))))))^2. - _Ilya Gutkovskiy_, May 10 2017

%p F:=n->add( n!*2^i/(1+i), i=0..n-1);

%t Table[n!Sum[2^j/(j+1),{j,0,n-1}],{n,0,30}] (* _Harvey P. Dale_, Jun 14 2017 *)

%o (Magma) [0] cat [Factorial(n)*(&+[2^j/(j+1):j in [0..n-1]]):n in [1..21]]; // _Marius A. Burtea_, Jul 12 2019

%o (PARI) a(n) = n!*sum(j=0, n-1, 2^j/(j+1)); \\ _Michel Marcus_, Jul 12 2019

%Y Row sums of A126671.

%K nonn

%O 0,3

%A _N. J. A. Sloane_ and Carlo Wood (carlo(AT)alinoe.com), Feb 13 2007