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a(n) = Sum_{k=0..n} 2^(n-k)*p(k), where p(k) is the partition function A000041.
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%I #17 Dec 03 2019 15:57:47

%S 1,3,8,19,43,93,197,409,840,1710,3462,6980,14037,28175,56485,113146,

%T 226523,453343,907071,1814632,3629891,7260574,14522150,29045555,

%U 58092685,116187328,232377092,464757194,929518106,1859040777,3718087158,7436181158,14872370665

%N a(n) = Sum_{k=0..n} 2^(n-k)*p(k), where p(k) is the partition function A000041.

%C In general, Sum_{k=0..n} (m^(n-k) * p(k)) ~ m^n / QPochhammer[1/m, 1/m], for m > 1.

%H Alois P. Heinz, <a href="/A259401/b259401.txt">Table of n, a(n) for n = 0..3320</a>

%F a(n) ~ c * 2^n, where c = 1/A048651 = 1/QPochhammer[1/2, 1/2] = 3.462746619455...

%F G.f.: (1/(1 - 2*x)) * Product_{k>=1} 1/(1 - x^k). - _Ilya Gutkovskiy_, Dec 03 2019

%p a:= proc(n) option remember; `if`(n<0, 0,

%p 2*a(n-1)+combinat[numbpart](n))

%p end:

%p seq(a(n), n=0..32); # _Alois P. Heinz_, Dec 03 2019

%t Table[Sum[2^(n-k)*PartitionsP[k],{k,0,n}],{n,0,50}]

%o (PARI) a(n) = sum(k=0, n, 2^(n-k)*numbpart(k)); \\ _Michel Marcus_, Dec 03 2019

%Y Cf. A000041, A048651, A090764, A259400, A292746.

%K nonn

%O 0,2

%A _Vaclav Kotesovec_, Jun 26 2015