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A067057
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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.
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1
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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
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OFFSET
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1,2
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COMMENTS
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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).
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LINKS
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FORMULA
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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]
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EXAMPLE
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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.
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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