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%I #34 Nov 03 2023 22:32:36
%S 0,0,1,1,1,1,1,2,3,4,4,5,6,8,10,12,14,17,20,24,29,34,40,47,55,64,75,
%T 87,101,117,135,155,179,205,235,269,307,350,399,453,514,583,660,746,
%U 843,950,1070,1205,1354,1520,1705,1910,2138,2392
%N Number of 3's in partitions of n into distinct parts.
%H Vaclav Kotesovec, <a href="/A015737/b015737.txt">Table of n, a(n) for n = 1..10000</a> (first 70 terms from Vincenzo Librandi)
%F G.f.: (x^3/(1 + x^3)) * Product_{j >= 1} (1 + x^j). - _Emeric Deutsch_, Apr 17 2006
%F Corresponding g.f. for "number of k's" is (x^k/(1 + x^k)) * Product_{j >= 1} (1 + x^j). - _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) = 3 because in the eight 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], only three contain 3.
%p g:=x^3*product(1+x^j,j=1..60)/(1+x^3): gser:=series(g,x=0,57): seq(coeff(gser,x,n),n=1..54); # _Emeric Deutsch_, Apr 17 2006
%t Table[Count[Select[IntegerPartitions[n],Length[Union[#]] == Length[#] &], _?(MemberQ[#, 3] &)], {n, 60}] (* _Harvey P. Dale_, Aug 19 2011 *)
%t nmax = 100; Rest[CoefficientList[Series[x^3/(1 + x^3) * Product[1 + x^k, {k, 1, nmax}], {x, 0, nmax}], x]] (* _Vaclav Kotesovec_, Oct 30 2015 *)
%Y Cf. A000009.
%K nonn
%O 1,8
%A _Clark Kimberling_
%E Example clarified by _Harvey P. Dale_, Aug 19 2011
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