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
Keywords
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)
Programs
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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`(i
n, 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 *)
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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
Comments