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A104456
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Number of ways of partitioning the integers {1,2,..,4n-1} into two (unordered) sets such that the sums of parts are equal in both sets (parts in one of the sets hence sum up to n*(4n-1)). Number of solutions to {1 +- 2 +- 3+ ... +- 4n-1 = 0}.
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3
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1, 4, 35, 361, 4110, 49910, 632602, 8273610, 110826888, 1512776590, 20965992017, 294245741167, 4173319332859, 59723919552183, 861331863890066, 12505857230438737, 182650875111521033, 2681644149792639400, 39555354718945873299, 585903163431438401072
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OFFSET
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1,2
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LINKS
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Alois P. Heinz, Table of n, a(n) for n = 1..250
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FORMULA
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a(n) is half the coefficient of q^(n*(4n - 1)) in product('1 + x^j', 'j'=1..4*n-1) for n >= 1.
n>=1, a(n)=(1/Pi)*2^(4n-1)*J(4n-1) where J(n) = integral(t=0, Pi/2, cos(t)cos(2t)...cos(nt)dt) - Benoit Cloitre, Sep 24 2006
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EXAMPLE
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a(2) = 4 since there are 4 ways of partitioning {1,2,3,4,5,6,7} into two sets of equal sum, namely {{1,2,5,6}, {3,4,7}}, {{1,3,4,6}, {2,5,7}}, {{2,3,4,5}, {1,6,7}}, {{1,2,4,7}, {3,5,6}}.
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MAPLE
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b:= proc(n, i) option remember; local m;
m:= i*(i+1)/2;
`if` (n>m, 0, `if` (n=m, 1, b(abs(n-i), i-1) +b(n+i, i-1)))
end:
a:= n-> b(4*n-1, 4*n-2):
seq (a(n), n=1..30); # Alois P. Heinz, Nov 01 2011
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MATHEMATICA
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Table[CoefficientList[Product[1 + x^j, {j, 1, 4n - 1}], x][[n*(4n - 1) + 1]]/2, {n, 1, 20}]
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CROSSREFS
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Cf. A060005. a(n)=A058377(4n-1). Also a(n) = A123117(n)/2.
Sequence in context: A188527 A026304 A215541 * A123224 A160887 A183878
Adjacent sequences: A104453 A104454 A104455 * A104457 A104458 A104459
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KEYWORD
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nonn
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AUTHOR
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Yiu Tung Poon (ytpoon(AT)iastate.edu) and Chun Chor Litwin Cheng (cccheng(AT)ied.edu.hk), Mar 08 2005
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STATUS
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approved
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