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%I #13 May 08 2020 12:45:43
%S 1,1,2,3,5,8,12,17,25,35,51,69,96,129,175,235,312,410,539,700,913,
%T 1173,1508,1923,2450,3105,3920,4926,6177,7710,9614,11923,14766,18218,
%U 22435,27550,33750,41231,50278,61150,74259,89932,108744,131193,158025,189979,227998,273125,326692
%N Expansion of Product_{k>=1} 1/((1 - x^k)*(1 - x^(5*k))).
%C Convolution of A000712 and A145466.
%C Convolution inverse of A030202.
%C Number of partitions of n if there are 2 kinds of parts that are multiples of 5.
%H Zakir Ahmed, Nayandeep Deka Baruah, Manosij Ghosh Dastidar, <a href="https://doi.org/10.1016/j.jnt.2015.05.002">New congruences modulo 5 for the number of 2-color partitions</a>, Journal of Number Theory, Volume 157, December 2015, Pages 184-198.
%H <a href="/index/Par#part">Index entries for sequences related to partitions</a>
%F G.f.: exp(Sum_{k>=1} x^k*(1 + x^k + x^(2*k) + x^(3*k) + 2*x^(4 k))/(k*(1 - x^(5*k)))).
%F a(n) ~ exp(2*Pi*sqrt(n/5)) / (4 * 5^(1/4) * n^(5/4)). - _Vaclav Kotesovec_, Aug 14 2018
%e a(5) = 8 because we have [5], [5'], [4, 1], [3, 2], [3, 1, 1], [2, 2, 1], [2, 1, 1, 1] and [1, 1, 1, 1, 1].
%p a:=series(mul(1/((1-x^k)*(1-x^(5*k))),k=1..55),x=0,49): seq(coeff(a,x,n),n=0..48); # _Paolo P. Lava_, Apr 02 2019
%t nmax = 48; CoefficientList[Series[Product[1/((1 - x^k) (1 - x^(5 k))), {k, 1, nmax}], {x, 0, nmax}], x]
%t nmax = 48; CoefficientList[Series[1/(QPochhammer[x] QPochhammer[x^5]), {x, 0, nmax}], x]
%t nmax = 48; CoefficientList[Series[Exp[Sum[x^k (1 + x^k + x^(2 k) + x^(3 k) + 2 x^(4 k))/(k (1 - x^(5 k))), {k, 1, nmax}]], {x, 0, nmax}], x]
%t Table[Sum[PartitionsP[k] PartitionsP[n - 5 k], {k, 0, n/5}], {n, 0, 48}]
%Y Cf. A000712, A002513, A030202, A030205, A145466, A318026, A318027.
%K nonn
%O 0,3
%A _Ilya Gutkovskiy_, Aug 13 2018