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A100405

Number of partitions of n where every part appears more than two times.

1, 0, 0, 1, 1, 1, 2, 1, 2, 3, 3, 3, 7, 5, 6, 11, 10, 10, 17, 15, 20, 26, 25, 29, 44, 41, 47, 63, 67, 72, 99, 97, 114, 143, 148, 168, 216, 216, 248, 306, 328, 358, 443, 462, 527, 629, 665, 739, 898, 936, 1055, 1238, 1330, 1465, 1727, 1837, 2055, 2366, 2543, 2808, 3274

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OFFSET

0,7

LINKS

Alois P. Heinz, Table of n, a(n) for n = 0..10000 (terms n = 1..5000 from Vaclav Kotesovec)

FORMULA

G.f.: Product_{k>0} (1+x^(3*k)/(1-x^k)). More generally, g.f. for number of partitions of n where every part appears more than m times is Product_{k>0} (1+x^((m+1)*k)/(1-x^k)).

a(n) ~ sqrt(Pi^2 + 6*c) * exp(sqrt((2*Pi^2/3 + 4*c)*n)) / (4*sqrt(3)*Pi*n), where c = Integral_{0..infinity} log(1 - exp(-x) + exp(-3*x)) dx = -0.77271248407593487127235205445116662610863126869049971822566... . - Vaclav Kotesovec, Jan 05 2016

EXAMPLE

a(6)=2 because we have [2,2,2] and [1,1,1,1,1,1].

MAPLE

G:=product((1+x^(3*k)/(1-x^k)), k=1..30): Gser:=series(G, x=0, 80): seq(coeff(Gser, x, n), n=0..70); # Emeric Deutsch, Aug 06 2005

# Alternative:

b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,

add(b(n-i*j, i-1), j=[0, $3..iquo(n, i)])))

end:

a:= n-> b(n$2):

seq(a(n), n=0..70); # Alois P. Heinz, Aug 20 2019

MATHEMATICA

nmax = 100; Rest[CoefficientList[Series[Product[1 + x^(3*k)/(1-x^k), {k, 1, nmax}], {x, 0, nmax}], x]] (* Vaclav Kotesovec, Nov 28 2015 *)