|
|
A085811
|
|
Number of partitions of n including 3, but not 1.
|
|
3
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,7
|
|
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).
|
|
LINKS
|
|
|
FORMULA
|
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
|
|
EXAMPLE
|
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
|
|
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: *)
|
|
PROG
|
(PARI) x='x+O('x^66); /* about that many terms */
v=Vec((x^3*(1-x)/eta(x))) /* Joerg Arndt, Feb 03 2012 */
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|