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%I #25 Sep 20 2014 02:46:37
%S 1,1,2,2,4,6,7,10,14,20,24,32,40,54,69,86,106,135,165,206,256,311,378,
%T 460,555,670,808,970,1156,1380,1638,1938,2296,2706,3188,3752,4390,
%U 5136,6000,6990,8128,9444,10944,12672,14659,16904,19476,22420,25753,29550,33873,38759,44306
%N Coefficients of the sum 1+ x/(1-x) + x^2/(1-x^2) + x^3/ ( (1-x)(1-x^2)) + x^4/ ( (1-x)(1-x^3) ) + x^5/ ( (1-x)(1-x^4) ) + x^5 /((1-x^2)(1-x^3)) + x^6/ ( (1-x)(1-x^2)(1-x^3)) + ...
%C For a given positive integer, n, let S_n be the set of partitions of n into distinct parts where the number of parts is maximal for that n. For example, for n=6, the set S_6 consists of just one such partition S_6={1,2,3}. Similarly, for n=7, S_7={1,2,4}, But for n=8, S_8 will contain two partitions S_8= { {1,2,5}, {1,3,4} }. Now form the sum whose general term is x^n divided by the product (1-x^(p_1))...(1-x^(p_i)) where the p's come from the partitions in S_n. The sequence is the sequence of coefficients of this sum.
%C This sequence is an upper bound for A098859.
%C The initial values of the sequence were verified by several members of the Sequence Fans mailing list.
%C |S_n| = A144328(n) (if the offset of the latter is changed to 0). - _Benoit Jubin_, Dec 13 2010.
%t <<Combinatorica`
%t f[set_]:=x^Total[set]/Product[1-x^set[[i]],{i,1,Length[set]}]
%t gf=1;
%t For[n=1,n<=20,n++,Print["n= ",n];
%t For[i=1,i<=Length[Partitions[n]],i++,
%t t=Tally[Partitions[n][[i]]];
%t sum=0;
%t For[j=1,j<=Length[t],j++,
%t If[t[[j]][[2]]==1,sum++]];
%t If[sum==Length[t]
%t &&Length[t]==Floor[(-1+Sqrt[1+8 n])/2],
%t gf=gf+f[Partitions[n][[i]]];Print[CoefficientList[Series[gf,{x,0,n}],x]];
%t ]]]
%Y Cf. A098859, A144328.
%K nonn
%O 0,3
%A _David S. Newman_, Dec 27 2010
%E More terms from _Alois P. Heinz_, Jun 25 2011
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