Complete partitions

franktaw at netscape.net franktaw at netscape.net
Thu Mar 22 23:01:43 CET 2007

```I just sent in the following comment:

%I A126796
%S A126796
1,1,2,2,4,5,8,10,16,20,31,39,55,71,100,125,173,218,291,366,483,600,784,97
1,1244,
1538,1957,2395,3023,3693,4605,5604,6942,8397,10347,12471,15235,18309,2226
7,26619,
32219,38414,46216,54941,65838,77958,93076,109908,130615,153855,182248,213
961,
252631,295913,348145,406826,477288,556230,650852,756881
%N A126796 Number of complete partitions of n.
%C A126796 A partition is complete iff each part is no more than 1 more
than the sum of all smaller parts.
%o A126796 (PARI)
T(n,k)=if(k<=1,1,if(n<2*k-1,T(n,floor((n+1)/2)),T(n,k-1)+T(n-k,k)))
a(n)=T(n,floor((n+1)/2)) /* If modified to save earlier results, this
would be efficient. */
%O A126796 1
%K A126796 ,nonn,
%A A126796 Franklin T. Adams-Watters (FrankTAW at Netscape.net), Mar 22
2007

It seems to me that, based on this, it should be possible to find a
generating function for this sequence.  I don't see how to, however.
Can somebody else on this list do it?

(The T(n,k) in the PARI program are the number of complete partitions
of n whose largest part is <= k.  Note the similarity to the triangle
of partition numbers, A008284.)