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%I #6 Aug 03 2014 14:27:18
%S 0,1,2,3,5,7,11,16,23,33,46,63,86,116,153,203,265,345,444,571,727,925,
%T 1166,1468,1836,2293,2845,3525,4345,5347,6550,8011,9758,11867,14380,
%U 17399,20984,25269,30341,36376,43500,51943,61877,73608,87373,103571
%N Number of partitions of n such that largest part k occurs at most floor(k/2) times.
%C Also number of partitions of n such that if the number of parts is k, then the smallest part is at most floor(k/2). Example: a(8)=16 because we have [7,1],[6,1,1],[5,2,1],[4,3,1],[5,1,1,1],[4,2,1,1],[3,3,1,1],[3,2,2,1],[2,2,2,2],[4,1,1,1,1],[3,2,1,1,1],[2,2,2,1,1],[3,1,1,1,1,1],[2,2,1,1,1,1],[2,1,1,1,1,1,1] and [1,1,1,1,1,1,1,1].
%F G.f.=sum(x^k*(1-x^(k(floor(k/2))))/product(1-x^j, j=1..k), k=1..infinity).
%e a(8)=16 because we have [8],[7,1],[6,2],[6,1,1],[5,3],[5,2,1],[5,1,1,1],[4,4],[4,3,1],[4,2,2],[4,2,1,1],[4,1,1,1,1],[3,2,2,1],[3,2,1,1,1],[3,1,1,1,1,1] and [2,1,1,1,1,1,1].
%p g:=sum(x^k*(1-x^(k*(floor(k/2))))/product(1-x^j,j=1..k),k=1..85): gser:=series(g,x=0,55): seq(coeff(gser,x,n),n=1..50);
%t z=55 ; q[n_] := q[n] = IntegerPartitions[n]; t[p_] := Length[p];
%t Table[Count[q[n], p_ /; 2 Min[p] <= t[p]], {n,z}] (* _Clark Kimberling_, Feb 15 2014 *)
%Y Cf. A118082, A118083.
%Y Cf. A237758, A237757, A237799, A237800. - _Clark Kimberling_, Feb 15 2014
%K nonn
%O 1,3
%A _Emeric Deutsch_, Apr 12 2006
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