A104456 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}.

1, 4, 35, 361, 4110, 49910, 632602, 8273610, 110826888, 1512776590, 20965992017, 294245741167, 4173319332859, 59723919552183, 861331863890066, 12505857230438737, 182650875111521033, 2681644149792639400, 39555354718945873299, 585903163431438401072

Offset

1,2

Links

Alois P. Heinz and Ray Chandler, Table of n, a(n) for n = 1..835 (terms < 10^1000, first 250 terms from Alois P. Heinz)

Formula

a(n) = A058377(4n-1). - N. J. A. Sloane, Jan 24 2006

a(n) is half the coefficient of q^(n*(4n - 1)) in the product('1 + x^j', 'j'=1..4*n-1), for n >= 1. - N. J. A. Sloane, Feb 24 2006

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), n>=1. - Benoit Cloitre, Sep 24 2006

a(n) = A123117(n)/2. - N. J. A. Sloane, Jan 09 2009

Example

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}}.
G.f. = x + 4*x^2 + 35*x^3 + 361*x^4 + 4110*x^5 + 49910*x^6 + ...

Maple

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

Mathematica

Table[CoefficientList[Product[1 + x^j, {j, 1, 4n - 1}], x][[n*(4n - 1) + 1]]/2, {n, 20}]

Crossrefs

Cf. A060005.

Sequence in context: A292190 A026304 A215541 * A277868 A305905 A324314

Adjacent sequences: A104453 A104454 A104455 * A104457 A104458 A104459

Keyword

nonn

Author

Yiu Tung Poon (ytpoon(AT)iastate.edu) and Chun Chor Litwin Cheng (cccheng(AT)ied.edu.hk), Mar 08 2005

Status

approved