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A067057
Let A(n) = {1,2,3,...n}. Let B(r) and C(n-r) be two subsets of A(n) having r and n-r elements respectively, such that B(r) U C(n-r) = A(n) and B and C are disjoint; then a(n) = sum of the products of all combination sums of elements of B and C for r =1 to n-1.
1
0, 2, 22, 140, 680, 2800, 10304, 34944, 111360, 337920, 985600, 2782208, 7641088, 20500480, 53903360, 139264000, 354287616, 889061376, 2203975680, 5404098560, 13120307200, 31569477632, 75342282752, 178467635200, 419849830400
OFFSET
1,2
COMMENTS
In other words, consider the set N = {1,2,3,...,n}; let S and S' be subsets of N such that S union S' is N. Define prod(S) = ( sum of members of S)*( sum of members of S'); then a(n) = sum of all possible prod(S).
FORMULA
For n>1, all listed values are given by a(n)=(2^(n-2))*s(n+1, n-1), where the s(n+1, n-1) are Stirling numbers of the first kind (A000914). - John W. Layman, Jan 05 2002
Conjecture: G.f.:(-2*x*(x+1))/(2*x-1)^5 [From Maksym Voznyy (voznyy(AT)mail.ru), Jul 26 2009]
EXAMPLE
For n = 4, N = {1,2,3,4}, the 5 columns below give S sum(S) S' sum(S') prod(S):
{ } 0 {1,2,3,4} 10 0
{1) 1 {2,3,4} 9 9
(2) 2 {1,3,4} 8 16
{3} 3 {1,2,4} 7 21
{4} 4 {1,2,3} 6 24
{1,2} 3 {3,4} 7 21
{1,3} 4 {2,4} 6 24
{1,4} 5 {2,3} 5 25
Hence a(4) = 1*(2 + 3 + 4) + 2*(1 + 3 + 4) + 3*(1 + 2 + 4) + 4*(1 + 2 + 3) + (1 + 2)*(3 + 4) + (1 + 3)*(2 + 4) + (1 + 4)*(2 + 3) = 140.
PROG
(PARI) print1(0, ", "); LIMIT = 40; V = vector(LIMIT*(LIMIT + 1)/2); V[1] = 1; for (i = 2, LIMIT, forstep (j = i*(i - 1)/2, 1, -1, V[i + j] += V[j]); V[i]++; k = i*(i + 1)/2; s = sum(j = 1, (k - 1)\2, j*(k - j)*V[j]); if (!(k%2), s += k*k*V[k\2]/8); print1(s, ", ")); \\ David Wasserman, Dec 22 2004
CROSSREFS
Cf. A000914.
Sequence in context: A062180 A244719 A084399 * A202738 A082940 A286778
KEYWORD
nonn
AUTHOR
Amarnath Murthy, Jan 02 2002, May 31 2003
EXTENSIONS
More terms and formula from John W. Layman, Jan 05 2002
Further terms from David Wasserman, Dec 22 2004
Edited by N. J. A. Sloane, Nov 01 2008 at the suggestion of R. J. Mathar
STATUS
approved