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G.f.: Sum_{k>=1} x^(k^2*(k + 1)/2) / (1 - x^(k^2*(k + 1)/2)).
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%I #10 Jan 02 2024 02:48:23

%S 1,1,1,1,1,2,1,1,1,1,1,2,1,1,1,1,1,3,1,1,1,1,1,2,1,1,1,1,1,2,1,1,1,1,

%T 1,3,1,1,1,2,1,2,1,1,1,1,1,2,1,1,1,1,1,3,1,1,1,1,1,2,1,1,1,1,1,2,1,1,

%U 1,1,1,3,1,1,2,1,1,2,1,2,1,1,1,2,1,1,1,1,1,3,1,1,1,1,1,2,1,1,1,1

%N G.f.: Sum_{k>=1} x^(k^2*(k + 1)/2) / (1 - x^(k^2*(k + 1)/2)).

%C Number of pentagonal pyramidal numbers (A002411) dividing n.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PentagonalPyramidalNumber.html">Pentagonal Pyramidal Number</a>.

%F Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Pi^2/3 - 2 = A195055 - 2 = 1.289868... . - _Amiram Eldar_, Jan 02 2024

%t nmax = 100; CoefficientList[Series[Sum[x^(k^2 (k + 1)/2)/(1 - x^(k^2 (k + 1)/2)), {k, 1, nmax}], {x, 0, nmax}], x] // Rest

%Y Cf. A002411, A007862, A195055, A279495, A279496, A279497.

%K nonn

%O 1,6

%A _Ilya Gutkovskiy_, May 16 2020