%I #9 Oct 19 2017 03:14:16
%S 0,1,2,4,5,9,12,19,25,37,49,70,90,124,161,216,275,363,460,597,750,960,
%T 1199,1519,1881,2359,2909,3617,4430,5469,6666,8173,9912,12079,14586,
%U 17680,21252,25630,30695,36848,43956,52547,62469,74383,88132,104556
%N a(n) = number of partitions of n wherein the sum of the 1's is no more than the sum of the other parts.
%F a(n) = A000041(n)-A000041(floor((n-1)/2)). - _Vladeta Jovovic_, Jun 15 2003
%e a(5)=5 because 5 = 1+4 = 1+1+3 = 1+2+2 = 2+3. (1+1+1+2 and 1+1+1+1+1 have too many ones.)
%t << DiscreteMath`Combinatorica`; f[n_] := Block[{c = 0, k = 1, p = Partitions[n], l = PartitionsP[n]}, While[k < l, c1 = Count[p[[k]], 1]; If[ Plus @@ Take[ p[[k]], Length[ p[[k]]] - c1] >= c1, c++ ]; k++ ]; c]; Table[ f[n], {n, 1, 25}]
%o (PARI) mybinary(n,m)=local(u); u=binary(n); concat(vector(m-length(u),i,0),u); { for (n=2,16,y=vector(2^(n-1)); for (j=0,2^(n-1)-1,x=concat(mybinary(j,n-1),[1]); y[j+1]=vector(n); c=0; for (k=1,n,if (x[k]==1,y[j+1][k]=c+1; c=0,c++)); y[j+1]=vecsort(y[j+1])); y=vecsort(y,QQ=QQ,2);
%o for (i=1,2^(n-1)-1,z=1; if (type(y[i])=="t_VEC", while (y[i]==y[i+z],y[i+z]=0; z++))); sc=0; for(i=1,2^(n-1),if (type(y[i])=="t_VEC",os=0; rs=0; for (q=1,n,if (y[i][q]==1,os++,rs+=y[i][q])); if (os<=rs,sc++))); print1(sc",") )}
%K nonn
%O 1,3
%A _Jon Perry_, Jun 15 2003
%E More terms from _Vladeta Jovovic_ and _Don Reble_, Jun 15 2003