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%I #21 May 16 2020 21:00:53
%S 0,0,0,0,1,1,1,2,2,2,3,4,4,6,8,9,11,14,16,19,23,27,32,38,45,53,62,72,
%T 84,97,112,130,150,172,199,228,260,298,340,386,440,500,566,642,727,
%U 820,926,1044,1174,1321,1484,1664,1866,2090
%N Number of 5's in all the partitions of n into distinct parts.
%H Vaclav Kotesovec, <a href="/A015740/b015740.txt">Table of n, a(n) for n = 1..10000</a>
%F G.f.: x^5*product(j>=1, 1+x^j )/(1+x^5). - _Emeric Deutsch_, Apr 17 2006
%F Corresponding g.f. for "number of k's" is x^k/(1+x^k)*prod(n>=1, 1+x^n ). [_Joerg Arndt_, Feb 20 2014]
%F a(n) ~ exp(Pi*sqrt(n/3)) / (8*3^(1/4)*n^(3/4)). - _Vaclav Kotesovec_, Oct 30 2015
%e a(9)=2 because in the 8 (=A000009(9)) partitions of 9 into distinct parts, namely [9],[8,1],[7,2],[6,3],[6,2,1],[5,4],[5,3,1] and [4,3,2] we have altogether two parts equal to 5.
%p g:=x^5*product(1+x^j,j=1..60)/(1+x^5): gser:=series(g,x=0,57): seq(coeff(gser,x,n),n=1..54); # _Emeric Deutsch_, Apr 17 2006
%t nmax = 100; Rest[CoefficientList[Series[x^5/(1+x^5) * Product[1+x^k, {k, 1, nmax}], {x, 0, nmax}], x]] (* _Vaclav Kotesovec_, Oct 30 2015 *)
%t Table[Count[Flatten@Select[IntegerPartitions[n], DeleteDuplicates[#] == # &], 5], {n, 54}] (* _Robert Price_, May 16 2020 *)
%K nonn
%O 1,8
%A _Clark Kimberling_
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