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A116645
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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).
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11
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,4
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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] * ... [From Mats Granvik, Gary W. Adamson, Aug 07 2009]
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LINKS
| Alois P. Heinz, Table of n, a(n) for n = 0..1000
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FORMULA
| G.f.: product((1-x^(2j)+x^(3j))/(1-x^j),j=1..infinity).
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
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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).
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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);
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CROSSREFS
| Column 0 of A116644.
Cf. A000041, A007690, A116595, A183559, A183568.
Sequence in context: A123632 A039868 A015723 * A177739 A193744 A039872
Adjacent sequences: A116642 A116643 A116644 * A116646 A116647 A116648
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KEYWORD
| nonn
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AUTHOR
| Emeric Deutsch (deutsch(AT)duke.poly.edu) and Vladeta Jovovic (vladeta(AT)eunet.rs), Feb 20 2006
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