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A181411 a(n) = Sum_{k=0..n} C(n,k)*sigma(n+k) for n>=1.

%I A181411 #10 Mar 27 2022 03:24:41
%S A181411 4,18,55,150,379,915,2146,4934,11080,24833,54476,119091,259432,556700,
%T A181411 1195135,2561094,5428597,11488866,24350993,51296325,107427025,
%U A181411 225330244,472762497,985966379,2049357779,4267962522,8887535983,18431783744
%N A181411 a(n) = Sum_{k=0..n} C(n,k)*sigma(n+k) for n>=1.
%H A181411 Vaclav Kotesovec, <a href="/A181411/b181411.txt">Table of n, a(n) for n = 1..3000</a>
%H A181411 Vaclav Kotesovec, <a href="/A181411/a181411.jpg">Plot of a(n)/(n*2^n) for n = 1..10000</a>
%F A181411 Equals the logarithmic derivative of A181410.
%F A181411 Conjecture: a(n) ~ c * n * 2^n, where c = Pi^2/4 = A091476. - _Vaclav Kotesovec_, Oct 05 2020
%e A181411 L.g.f.: L(x) = 4*x + 18*x^2/2 + 55*x^3/3 + 150*x^4/4 + 379*x^5/5 +...
%e A181411 Exponentiation yields the g.f. of A181410:
%e A181411 exp(L(x)) = 1 + 4*x + 17*x^2 + 65*x^3 + 234*x^4 + 804*x^5 +...
%e A181411 The initial terms begin:
%e A181411 a(1) = 1*1 + 1*3 = 4;
%e A181411 a(2) = 1*3 + 2*4 + 1*7 = 18;
%e A181411 a(3) = 1*4 + 3*7 + 3*6 + 1*12 = 55;
%e A181411 a(4) = 1*7 + 4*6 + 6*12 + 4*8 + 1*15 = 150; ...
%t A181411 Table[Sum[Binomial[n,k] * DivisorSigma[1,n+k],{k,0,n}],{n,1,30}] (* _Vaclav Kotesovec_, Oct 05 2020 *)
%o A181411 (PARI) {a(n)=sum(k=0,n,binomial(m,k)*sigma(n+k))}
%Y A181411 Cf. A181410.
%K A181411 nonn
%O A181411 1,1
%A A181411 _Paul D. Hanna_, Oct 19 2010