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A116645
Number of partitions of n having no doubletons. By a doubleton in a partition we mean an occurrence of a part exactly twice (the partition [4,(3,3),2,2,2,(1,1)] of 18 has two doubletons, shown between parentheses).
12
1, 1, 1, 3, 3, 5, 8, 10, 13, 20, 26, 33, 46, 58, 75, 101, 125, 157, 206, 253, 317, 403, 494, 608, 760, 926, 1131, 1393, 1685, 2038, 2487, 2985, 3585, 4331, 5168, 6172, 7392, 8771, 10410, 12382, 14622, 17258, 20400, 23975, 28159, 33115, 38739, 45298, 53000
OFFSET
0,4
COMMENTS
Number of partitions of n having no part that appears exactly twice.
Infinite convolution product of [1,1,0,1,1,1,1,1,1,1] aerated n-1 times. I.e., [1,1,0,1,1,1,1,1,1,1] * [1,0,1,0,0,0,1,0,1,0] * [1,0,0,1,0,0,0,0,0,1] * ... - Mats Granvik, Gary W. Adamson, Aug 07 2009
LINKS
FORMULA
G.f.: Product_{j>=1} (1-x^(2j)+x^(3j))/(1-x^j).
G.f. for the number of partitions of n having no part that appears exactly m times is Product_{k>0} (1/(1-x^k)-x^(m*k)).
a(n) = A000041(n) - A183559(n) = A183568(n,0) - A183568(n,2). - Alois P. Heinz, Oct 09 2011
EXAMPLE
a(4) = 3 because we have [4],[3,1] and [1,1,1,1] (the partitions [2,2] and [2,1,1] do not qualify since each of them has a doubleton).
MAPLE
h:=product((1-x^(2*j)+x^(3*j))/(1-x^j), j=1..60): hser:=series(h, x=0, 60): seq(coeff(hser, x, n), n=0..56);
MATHEMATICA
nn=48; CoefficientList[Series[Product[1/(1-x^i)-x^(2i), {i, 1, nn}], {x, 0, nn}], x] (* Geoffrey Critzer, Sep 30 2013 *)
CROSSREFS
Column 0 of A116644.
Sequence in context: A015723 A333150 A342343 * A177739 A323581 A327731
KEYWORD
nonn
AUTHOR
STATUS
approved