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Search: id:A116928
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| A116928 |
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Number of 1's in all self-conjugate partitions of n. |
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+0
2
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| 1, 0, 1, 0, 2, 1, 3, 2, 4, 4, 6, 6, 8, 9, 11, 12, 15, 17, 20, 22, 26, 29, 34, 37, 43, 48, 55, 60, 69, 76, 86, 94, 106, 117, 131, 143, 160, 176, 195, 213, 236, 259, 285, 311, 342, 374, 410, 446, 488, 533, 581, 631, 688, 748, 813, 881, 957, 1038, 1125, 1216, 1317, 1425
(list; graph; listen)
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OFFSET
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1,5
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COMMENT
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a(n)=Sum(k*A116927(n,k), k>=0).
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FORMULA
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G.f.=x+sum(x^(k^2+2)/(1-x^2)/product(1-x^(2j), j=1..k), k=1..infinity).
a(n) = A096911(n)-(1+(-1)^n)/2, m>1. - Vladeta Jovovic (vladeta(AT)Eunet.yu), Feb 27 2006
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EXAMPLE
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a(12)=6 because the self-conjugate partitions of 12 are [6,2,1,1,1,1],[5,3,2,1,1], and [4,4,2,2], containing a total of six 1's.
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MAPLE
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f:=x+sum(x^(k^2+2)/(1-x^2)/product(1-x^(2*j), j=1..k), k=1..10): fser:=series(f, x=0, 70): seq(coeff(fser, x^n), n=1..67);
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CROSSREFS
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Cf. A116927.
Sequence in context: A029139 A100927 A001687 this_sequence A034391 A094173 A026272
Adjacent sequences: A116925 A116926 A116927 this_sequence A116929 A116930 A116931
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KEYWORD
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nonn
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AUTHOR
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Emeric Deutsch (deutsch(AT)duke.poly.edu), Feb 26 2006
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