A117989 Number of partitions of n such that the least part occurs at least twice.

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

Offset: 1

Author

Emeric Deutsch, Apr 08 2006

Keywords

• nonn

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

Examples

			a(5) = 3 because we have [3,1,1], [2,1,1,1] and [1,1,1,1,1].
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..1000
• Aritram Dhar, Proofs of Two Formulas of Vladeta Jovovic, arXiv:2112.07762 [math.CO], 2021.

Crossrefs

Cf. A096373, A000041, A056823.

Cf. A097364, A323111.

Programs

Haskell

a117989 n = a117989_list !! (n-1)
a117989_list = tail $ zipWith (-)
                      (map (* 2) a000041_list) $ tail a000041_list
-- Reinhard Zumkeller, Nov 12 2015

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);
A117989 := proc(n)
    2*combinat[numbpart](n)-combinat[numbpart](n+1) ;
end proc: # R. J. Mathar, May 19 2016

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 *)

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) = 2*A000041(n) - A000041(n+1). - Vladeta Jovovic, Jul 21 2006

a(n) = A056823(n+1) - 2*A056823(n). - Bob Selcoe, Apr 11 2014

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