|
Search: id:A117989
|
|
|
| A117989 |
|
Number of partitions of n such that the least part occurs at least twice. |
|
+0
2
|
|
| 0, 1, 1, 3, 3, 7, 8, 14, 18, 28, 35, 53, 67, 94, 121, 165, 209, 280, 353, 462, 582, 749, 935, 1192, 1480, 1862, 2302, 2871, 3526, 4366, 5335, 6555, 7976, 9737, 11789, 14317, 17259, 20845, 25032, 30093, 35992, 43087, 51347, 61216, 72710, 86362, 102235
(list; graph; listen)
|
|
|
OFFSET
|
1,4
|
|
|
COMMENT
|
More generally, g.f. for number of partitions of n such that the least part occurs at least m times is sum(x^(mk)/product(1-x^j, j=k..infinity), k=1..infinity). Also number of partitions of n such that if k is the largest part, then k>=2 and k-1 does not occur. Example: a(5)=3 because we have [5],[4,1], and [3,1,1].
Also number of partitions of 2*n such that the difference between greatest part and smallest part is n. - Vladeta Jovovic (vladeta(AT)Eunet.yu), May 09 2008
|
|
FORMULA
|
G.f.=sum(x^(2k)/product(1-x^j, j=k..infinity), k=1..infinity). G.f.=sum(x^k*(1-x^(k-1))/product(1-x^j, j=1..k), k=2..infinity).
a(n) = 2*A000041(n)-A000041(n+1). - Vladeta Jovovic (vladeta(AT)Eunet.yu), Jul 21 2006
|
|
EXAMPLE
|
a(5)=3 because we have [3,1,1],[2,1,1,1], and [1,1,1,1,1].
|
|
MAPLE
|
g:=sum(x^k*(1-x^(k-1))/product(1-x^j, j=1..k), k=2..70): gser:=series(g, x=0, 55): seq(coeff(gser, x, n), n=1..50);
|
|
CROSSREFS
|
Cf. A096373.
Cf. A097364.
Sequence in context: A021300 A121527 A021886 this_sequence A086543 A110618 A108046
Adjacent sequences: A117986 A117987 A117988 this_sequence A117990 A117991 A117992
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Emeric Deutsch (deutsch(AT)duke.poly.edu), Apr 08 2006
|
|
|
Search completed in 0.002 seconds
|
