OFFSET
1,2
COMMENTS
Also, the number of partitions of n such that (greatest part) <= 2*(number of parts); hence, the number of partitions of n such that (rank + greatest part) >= 0.
LINKS
Seiichi Manyama, Table of n, a(n) for n = 1..1000
FORMULA
G.f.: Sum_{n>=1} x^n * Product_{k=1..n} (1 - x^(2*n+k-1))/(1 - x^k). - Paul D. Hanna, Aug 03 2015
EXAMPLE
a(6) = 9 counts all of the 11 partitions of 6 except these: 21111, 111111.
MATHEMATICA
z = 50; Table[Count[IntegerPartitions[n], p_ /; 2 Max[p] >= Length[p]], {n, z}]
PROG
(PARI) {a(n) = my(A); A = sum(m=0, n, x^m*prod(k=1, m, (1-x^(2*m+k-1))/(1-x^k +x*O(x^n)))); polcoeff(A, n)}
for(n=1, 60, print1(a(n), ", ")) \\ Paul D. Hanna, Aug 03 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Feb 13 2014
STATUS
approved