|
| |
|
|
A117989
|
|
Number of partitions of n such that the least part occurs at least twice.
|
|
24
|
|
|
|
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;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,4
|
|
|
COMMENTS
|
More generally, the g.f. for the 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, the 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, the number of partitions of 2n such that the difference between greatest part and smallest part is n. - Vladeta Jovovic, May 09 2008
|
|
|
LINKS
|
|
|
|
FORMULA
|
G.f.: sum(k>=1, x^(2*k)/prod(j>=k, 1-x^j ) ).
G.f.: sum(k>=1, x^k*(1-x^(k-1))/prod(j=1..k, 1-x^j ) ).
a(n) ~ exp(Pi*sqrt(2*n/3)) / (4*sqrt(3)*n) * (1 - (sqrt(3/2)/Pi + 25*Pi/(24*sqrt(6))) / sqrt(n)). - Vaclav Kotesovec, Nov 03 2020
|
|
|
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);
2*combinat[numbpart](n)-combinat[numbpart](n+1) ;
|
|
|
MATHEMATICA
|
Table[Length[Select[IntegerPartitions[n], Count[#, Min[#]]>1&]], {n, 50}] (* Harvey P. Dale, Apr 23 2011 *)
max = 48; Sum[x^(2*k)/Product[1 - x^j, {j, k, Infinity}], {k, 1, Ceiling[ max/2]}] + O[x]^max // CoefficientList[#, x]& // Rest (* Jean-François Alcover, Sep 11 2017 *)
|
|
|
PROG
|
(Haskell)
a117989 n = a117989_list !! (n-1)
a117989_list = tail $ zipWith (-)
(map (* 2) a000041_list) $ tail a000041_list
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
