A101198 Number of partitions of n with rank 1 (the rank of a partition is the largest part minus the number of parts).
0, 1, 0, 1, 1, 2, 1, 3, 3, 5, 5, 8, 8, 13, 14, 20, 23, 31, 35, 48, 55, 72, 84, 108, 126, 160, 187, 233, 275, 340, 398, 489, 574, 697, 819, 988, 1158, 1390, 1627, 1941, 2271, 2696, 3145, 3721, 4335, 5104, 5938, 6967, 8088, 9462, 10964, 12783
Offset: 1
Keywords
Examples
a(6)=2 because the 11 partitions 6,51,42,411,33,321,3111,222,2211,21111,111111 have ranks 5,3,2,1,1,0,-1,-1,-2,-3,-5, respectively.
References
- George E. Andrews, The Theory of Partitions, Addison-Wesley, Reading, Mass., 1976.
Links
- Seiichi Manyama, Table of n, a(n) for n = 1..10000
Programs
-
Maple
with(combinat): for n from 1 to 35 do P:=partition(n): c:=0: for j from 1 to nops(P) do if P[j][nops(P[j])]-nops(P[j])=1 then c:=c+1 else c:=c fi od: a[n]:=c: od: seq(a[n],n=1..35);
-
Mathematica
Table[Count[IntegerPartitions[n],?(Max[#]-Length[#]==1&)],{n,60}] (* _Harvey P. Dale, Nov 29 2014 *)
Formula
G.f. for the number of partitions of n with rank r is Sum((-1)^k*x^(r*k)*(x^((3*k^2+k)/2)-x^((3*k^2-k)/2)), k=1..infinity)/Product(1-x^k, k=1..infinity). - Vladeta Jovovic, Dec 20 2004
Also Sum(x^(2*n+r+1)*Product((1-x^(2*n+r+1-k))/(1-x^k),k=1..n),n=0..infinity). - Vladeta Jovovic, May 05 2008
a(n) ~ Pi * exp(Pi*sqrt(2*n/3)) / (3 * 2^(9/2) * n^(3/2)). - Vaclav Kotesovec, May 26 2023
Comments