%I #13 May 08 2020 12:45:10
%S 1,1,2,4,6,9,16,22,33,50,70,98,143,193,266,368,493,659,892,1170,1543,
%T 2035,2642,3422,4448,5694,7294,9334,11839,14982,18968,23812,29868,
%U 37410,46598,57924,71953,88913,109728,135212,165991,203407,248986,303706,369939,449967,545820,661038,799629
%N Expansion of Product_{k>=1} 1/((1 - x^k)*(1 - x^(3*k))).
%C Convolution of A000041 and A035377.
%C Convolution of A000712 and A137569.
%C Convolution inverse of A030203.
%C Number of partitions of n if there are 2 kinds of parts that are multiples of 3.
%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 + 2*x^(2*k))/(k*(1 - x^(3*k)))).
%F a(n) ~ exp(2*sqrt(2*n)*Pi/3) / (3 * 2^(5/4) * n^(5/4)). - _Vaclav Kotesovec_, Aug 14 2018
%e a(4) = 6 because we have [4], [3, 1], [3', 1], [2, 2], [2, 1, 1] and [1, 1, 1, 1].
%p a:=series(mul(1/((1-x^k)*(1-x^(3*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^(3 k))), {k, 1, nmax}], {x, 0, nmax}], x]
%t nmax = 48; CoefficientList[Series[1/(QPochhammer[x] QPochhammer[x^3]), {x, 0, nmax}], x]
%t nmax = 48; CoefficientList[Series[Exp[Sum[x^k (1 + x^k + 2 x^(2 k))/(k (1 - x^(3 k))), {k, 1, nmax}]], {x, 0, nmax}], x]
%t Table[Sum[PartitionsP[k] PartitionsP[n - 3 k], {k, 0, n/3}], {n, 0, 48}]
%Y Cf. A000041, A000712, A002513, A030203, A030206, A035377, A112175, A137569, A318027, A318028.
%K nonn
%O 0,3
%A _Ilya Gutkovskiy_, Aug 13 2018