A118096 Number of partitions of n such that the largest part is twice the smallest part.

0, 0, 1, 1, 2, 3, 3, 4, 6, 6, 6, 10, 9, 11, 13, 14, 15, 20, 18, 23, 25, 27, 27, 37, 35, 39, 43, 48, 49, 61, 57, 68, 72, 78, 81, 97, 95, 107, 114, 127, 128, 150, 148, 168, 179, 191, 198, 229, 230, 254, 266, 291, 300, 338, 344, 379, 398, 427, 444, 498, 505, 550, 580, 625

Offset: 1

Author

Emeric Deutsch, Apr 12 2006

Keywords

• nonn

Comments

Also number of partitions of n such that if the largest part occurs k times, then the number of parts is 2k. Example: a(8)=4 because we have [7,1], [6,2], [5,3] and [3,3,1,1].

Examples

			a(8)=4 because we have [4,2,2], [2,2,2,1,1], [2,2,1,1,1,1] and [2,1,1,1,1,1,1].
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Alois P. Heinz)

Crossrefs

Cf. A237825, A237826, A237827.

Cf. A008483, A117086, A238479, A350893.

Programs

Maple

g:=sum(x^(3*k)/product(1-x^j,j=k..2*k),k=1..30): gser:=series(g,x=0,75): seq(coeff(gser,x,n),n=1..70);
# second Maple program:
b:= proc(n, i, t) option remember: `if`(n=0, 1, `if`(in, 0, b(n-i, i, t))))
    end:
a:= n-> add(b(n-3*j, 2*j, j), j=1..n/3):
seq(a(n), n=1..64);  # Alois P. Heinz, Sep 04 2017

Mathematica

Table[Count[IntegerPartitions[n], p_ /; 2 Min[p] = = Max[p]], {n, 40}] (* Clark Kimberling, Feb 16 2014 *)
(* Second program: *)
b[n_, i_, t_] := b[n, i, t] = If[n == 0, 1, If[i < t, 0,
     b[n, i - 1, t] + If[i > n, 0, b[n - i, i, t]]]];
a[n_] := Sum[b[n - 3j, 2j, j], {j, 1, n/3}];
Array[a, 64] (* Jean-François Alcover, Jun 04 2021, after Alois P. Heinz *)
(* Third program: *)
nmax = 100; p = 1; s = 0; Do[p = Simplify[p*(1 - x^(2*k - 1))*(1 - x^(2*k))/(1 - x^k)]; p = Normal[p + O[x]^(nmax+1)]; s += x^(3*k)/(1 - x^k)/p;, {k, 1, nmax}]; Rest[CoefficientList[Series[s, {x, 0, nmax}], x]] (* Vaclav Kotesovec, Jun 16 2025 *)

PARI

my(N=70, x='x+O('x^N)); concat([0, 0], Vec(sum(k=1, N, x^(3*k)/prod(j=k, 2*k, 1-x^j)))) \\ Seiichi Manyama, May 14 2023

Formula

G.f.: Sum_{k>=1} x^(3*k)/Product_{j=k..2*k} (1-x^j).

a(n) ~ exp(Pi*sqrt(2*n/15)) / (5^(1/4)*sqrt(2*phi*n)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Jun 13 2025