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

1,6

Comments

Column k=1 in the triangle A063995.

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

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

Example

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.

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 *)

Crossrefs

Cf. A000041, A063995.

Cf. A101198-A101200, A101707-A101709.

Sequence in context: A074500 A107237 A070047 * A357643 A034394 A058689

Adjacent sequences: A101195 A101196 A101197 * A101199 A101200 A101201

Keyword

nonn

Author

Emeric Deutsch, Dec 12 2004

Status

approved