A100824 Number of partitions of n with at most one odd part.
1, 1, 1, 2, 2, 4, 3, 7, 5, 12, 7, 19, 11, 30, 15, 45, 22, 67, 30, 97, 42, 139, 56, 195, 77, 272, 101, 373, 135, 508, 176, 684, 231, 915, 297, 1212, 385, 1597, 490, 2087, 627, 2714, 792, 3506, 1002, 4508, 1255, 5763, 1575, 7338, 1958, 9296, 2436, 11732, 3010, 14742
Offset: 0
Examples
From _Gus Wiseman_, Jan 21 2022: (Start) The a(1) = 1 through a(9) = 12 partitions with at most one odd part: (1) (2) (3) (4) (5) (6) (7) (8) (9) (21) (22) (32) (42) (43) (44) (54) (41) (222) (52) (62) (63) (221) (61) (422) (72) (322) (2222) (81) (421) (432) (2221) (441) (522) (621) (3222) (4221) (22221) (End)
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1000
Crossrefs
Programs
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Maple
seq(coeff(convert(series((1+x/(1-x^2))/mul(1-x^(2*i),i=1..100),x,100),polynom),x,n),n=0..60); (C. Ronaldo)
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Mathematica
nmax = 50; CoefficientList[Series[(1+x/(1-x^2)) * Product[1/(1-x^(2*k)), {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Mar 07 2016 *) Table[Length[Select[IntegerPartitions[n],Count[#,?OddQ]<=1&]],{n,0,30}] (* _Gus Wiseman, Jan 21 2022 *)
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PARI
a(n) = if(n%2==0, numbpart(n/2), sum(i=1, (n+1)\2, numbpart((n-2*i+1)\2))) \\ David A. Corneth, Jan 23 2022
Formula
G.f.: (1+x/(1-x^2))/Product(1-x^(2*i), i=1..infinity). More generally, g.f. for number of partitions of n with at most k odd parts is (1+Sum(x^i/Product(1-x^(2*j), j=1..i), i=1..k))/Product(1-x^(2*i), i=1..infinity).
a(n) ~ exp(sqrt(n/3)*Pi) / (2*sqrt(3)*n) if n is even and a(n) ~ exp(sqrt(n/3)*Pi) / (2*Pi*sqrt(n)) if n is odd. - Vaclav Kotesovec, Mar 07 2016
Extensions
More terms from C. Ronaldo (aga_new_ac(AT)hotmail.com), Jan 19 2005
Comments