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L.g.f.: -log(Product_{k>=1} (1 - k*x^k)^k) = Sum_{n>=1} a(n)*x^n/n.
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%I #28 Jun 24 2019 20:10:06

%S 1,9,28,81,126,330,344,833,973,1754,1332,5034,2198,5658,8688,13313,

%T 4914,28779,6860,54106,45752,33482,12168,254954,93751,78906,255880,

%U 505698,24390,1510700,29792,1671169,1791312,647114,2819544,12637371,50654,2282346,14779520,34058298,68922,68084220

%N L.g.f.: -log(Product_{k>=1} (1 - k*x^k)^k) = Sum_{n>=1} a(n)*x^n/n.

%H Seiichi Manyama, <a href="/A296601/b296601.txt">Table of n, a(n) for n = 1..5000</a>

%H <a href="/index/Su#sums_of_divisors">Index entries for sequences related to sums of divisors</a>

%F G.f.: Sum_{k>=1} k^3*x^k/(1 - k*x^k).

%F a(n) = Sum_{d|n} d^(n/d+2).

%F a(p) = p^3 + 1 where p is a prime.

%F From _Seiichi Manyama_, Jun 24 2019: (Start)

%F Suppose given two sequences f(n) and g(n), n>0, we define a new sequence a(n), n>0, by a(n) = Sum_{d|n} d*f(d)*g(d)^(n/d).

%F L.g.f.: -log(Product_{n>0} (1 - g(n)*x^n)^f(n)) = Sum_{n>0} a(n)*x^n/n. (See A266964.)

%F If we set f(n) = n and g(n) = n, we get this sequence. (End)

%e L.g.f.: L(x) = x + 9*x^2/2 + 28*x^3/3 + 81*x^4/4 + 126*x^5/5 + 330*x^6/6 + 344*x^7/7 + 833*x^8/8 + 973*x^9/9 + ...

%e exp(L(x)) = 1 + x + 5*x^2 + 14*x^3 + 42*x^4 + 103*x^5 + 289*x^6 + 690*x^7 + 1771*x^8 + 4206*x^9 + ... + A266941(n)*x^n + ...

%t nmax = 42; Rest[CoefficientList[Series[-Log[Product[(1 - k x^k)^k, {k, 1, nmax}]], {x, 0, nmax}], x] Range[0, nmax]]

%t nmax = 42; Rest[CoefficientList[Series[Sum[k^3 x^k/(1 - k x^k), {k, 1, nmax}], {x, 0, nmax}], x]]

%t a[n_] := Sum[d^(n/d + 2), {d, Divisors[n]}]; Table[a[n], {n, 1, 42}]

%o (PARI) N=66; x='x+O('x^N); Vec(x*deriv(-log(prod(k=1, N, (1-k*x^k)^k)))) \\ _Seiichi Manyama_, Jun 02 2019

%Y Column k=2 of A308502.

%Y Cf. A001157, A078308, A266941, A266964.

%K nonn

%O 1,2

%A _Ilya Gutkovskiy_, May 20 2018