1,4

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

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.

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

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

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

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 *)

(Haskell)

a117989 n = a117989_list !! (n-1)

a117989_list = tail $ zipWith (-)

(map (* 2) a000041_list) $ tail a000041_list

-- Reinhard Zumkeller, Nov 12 2015

Cf. A096373, A000041, A056823.

Cf. A097364, A323111.

Sequence in context: A218571 A218572 A218573 * A391794 A241642 A086543

Adjacent sequences: A117986 A117987 A117988 * A117990 A117991 A117992

Emeric Deutsch, Apr 08 2006

approved