A085811 Number of partitions of n including 3, but not 1.

0, 0, 1, 0, 1, 1, 2, 2, 4, 4, 7, 8, 12, 14, 21, 24, 34, 41, 55, 66, 88, 105, 137, 165, 210, 253, 320, 383, 478, 574, 708, 847, 1039, 1238, 1507, 1794, 2167, 2573, 3094, 3660, 4378, 5170, 6153, 7245, 8591, 10087, 11914, 13959, 16424, 19196, 22519, 26252, 30701

Offset: 1

Author

Jon Perry, Jul 25 2003

Keywords

• nonn

Comments

Related to the 'number of sums containing k' phenomena reported at link. Define P_k(n,j) to be the number of partitions of n with minimum part j and containing k, P_k(n) as the number of partitions of n that contain k as a part and P(n,j) as the number of partitions of k that have minimum part k, then: P_k(n)=sum{i=1,k-1,P_k(n-i,i)}+P(n-k,k) which (unproved) gives P(n-k). This sequence gives P_3(n,2). E.g. assume P_3(9)=11. P_3(10)=P_3(9,1)+P_3(8,2)+P(7,3)=11+2+2=15, where P(7,3) is given by A008483(7).

Examples

			a(3): 3
a(5): 2+3
a(6): 3+3
a(7): 2+2+3, 3+4
a(8): 2+3+3, 3+5
a(9): 2+3+4, 2+2+2+3, 3+3+3, 3+6
a(10): 2+3+5, 2+2+3+3, 3+7, 3+3+4
a(11): 2+2+3+4, 2+3+6, 2+2+2+2+3, 2+3+3+3, 3+4+4, 3+8, 3+3+5,
a(12): 2+2+2+3+3, 2+3+3+4, 2+3+7, 2+2+3+5, 3+9, 3+3+6, 3+4+5, 3+3+3+3
a(13): 2+2+2+2+2+3, 2+2+2+3+4, 2+2+3+6, 2+2+3+3+3, 2+3+4+4, 2+3+3+5,
2+3+8, 3+10, 3+3+7, 3+4+6, 3+5+5, 3+3+3+4
		

Links

• Andrew van den Hoeven, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A008483, A002865. Essentially the same as A002865.

Programs

Mathematica

f[n_] := Block[{c = 0, k = 1, m = PartitionsP[n], p = IntegerPartitions[n] }, While[k < m, If[ Count[ p[[k]], 3] > 0 && Count[ p[[k]], 1] == 0, c++ ]; k++ ]; c]; Table[ f[n], {n, 1, 53}]
(* second program: *)
CoefficientList[x^2*(1-x)/QPochhammer[x] + O[x]^60, x] (* Jean-François Alcover, Jan 22 2016, after Joerg Arndt *)

PARI

x='x+O('x^66); /* about that many terms */
v=Vec((x^3*(1-x)/eta(x)))  /* Joerg Arndt, Feb 03 2012 */

Formula

A002865(n) = a(n+3). - James Sellers, Dec 06 2005.

G.f.: x^3*(1-x)/prod(n>=1, 1-x^n). [Joerg Arndt, Feb 03 2012]

G.f.: x^2 + x^3*(1 - G(0))/(1-x) where G(k) = 1 - x/(1-x^(k+1))/(1-x/(x-1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 23 2013

Extensions

Edited, corrected and extended by Robert G. Wilson v

Typo in formula corrected by Andrew van den Hoeven, Nov 20 2014