A364910 Number of integer partitions of 2n whose distinct parts sum to n.
1, 1, 1, 3, 3, 4, 12, 11, 19, 23, 54, 55, 103, 115, 178, 289, 389, 507, 757, 970, 1343, 2033, 2579, 3481, 4840, 6312, 8317, 10998, 15459, 19334, 26368, 33480, 44709, 56838, 74878, 93369, 128109, 157024, 206471, 258357, 338085, 417530, 544263, 669388, 859570, 1082758, 1367068
Offset: 0
Keywords
Examples
The a(0) = 1 through a(7) = 11 partitions: () (11) (22) (33) (44) (55) (66) (77) (2211) (3311) (3322) (4422) (4433) (21111) (311111) (4411) (5511) (5522) (4111111) (33321) (6611) (42222) (442211) (322221) (4222211) (332211) (4421111) (3222111) (42221111) (3321111) (422111111) (32211111) (611111111) (51111111) (4211111111) (321111111) The a(0) = 1 through a(7) = 11 linear combinations: 0 1*1 1*2 1*3 1*4 1*5 1*6 1*7 0*2+3*1 0*3+4*1 0*4+5*1 0*4+3*2 0*6+7*1 1*2+1*1 1*3+1*1 1*3+1*2 0*5+6*1 1*4+1*3 1*4+1*1 1*4+1*2 1*5+1*2 1*5+1*1 1*6+1*1 0*3+0*2+6*1 0*4+0*2+7*1 0*3+1*2+4*1 0*4+1*2+5*1 0*3+2*2+2*1 0*4+2*2+3*1 0*3+3*2+0*1 0*4+3*2+1*1 1*3+0*2+3*1 1*4+0*2+3*1 1*3+1*2+1*1 1*4+1*2+1*1 2*3+0*2+0*1
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..500 (first 91 terms from David A. Corneth)
Crossrefs
The case with no zero coefficients is A000009.
Central diagonal of A116861.
A version based on Heinz numbers is A364906.
Using all partitions (not just strict) we get A364907.
Main diagonal of A364916.
Using strict partitions of any number from 1 to n gives A365002.
These partitions have ranks A365003.
Programs
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Mathematica
Table[Length[Select[IntegerPartitions[2n],Total[Union[#]]==n&]],{n,0,15}]
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PARI
a(n) = {my(res = 0); forpart(p = 2*n,s = Set(p); if(vecsum(s) == n, res++)); res} \\ David A. Corneth, Aug 20 2023
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Python
from sympy.utilities.iterables import partitions def A364910(n): return sum(1 for d in partitions(n<<1,k=n) if sum(set(d))==n) # Chai Wah Wu, Sep 13 2023
Extensions
More terms from David A. Corneth, Aug 20 2023
Comments