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A087127
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This table shows the sobalian coefficients of combinatorial formulae needed for generating the sequential sums of p-th powers of triangular numbers. The p-th row (p>=1) contains a(i,p) for i=1 to 2*p-1, where a(i,p) satisfies Sum_{i=1..n} C(i+1,2)^p = 3 * C(n+2,3) * Sum_{i=1..2*p-1} a(i,p) * C(n-1,i-1)/(i+2).
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26
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1, 1, 2, 1, 1, 8, 19, 18, 6, 1, 26, 163, 432, 564, 360, 90, 1, 80, 1135, 6354, 18078, 28800, 26100, 12600, 2520, 1, 242, 7291, 77400, 405060, 1210680, 2211570, 2520000, 1751400, 680400, 113400, 1, 728, 45199, 862218, 7667646, 38350080, 118848420
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,3
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FORMULA
| a(1, p) = 1, a(2, p) = 3^(p-1)-1, a(3, p) = 3^(p-1)*[2^(p-1)-2]+1, ..., a(2*p-3, p) = [ (6*p^4-20*p^3+21*p^2-7*p)*(2*p-4)! ]/[3*2^(p-1)], a(2*p-2, p) = [ (p^2-p)*(2*p-3)! ]/2^(p-2), a(2*p-1, p) = [ (p-1)*(2*p-3)! ]/2^(p-2).
a(i, p) = Sum_{k=1..[2*i+1+(-1)^(i-1)]/4} [ C(i-1, 2*k-2)*C(i-2*k+3, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+2, i-2*k)^(p-1) ]
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EXAMPLE
| Row 3 contains 1,8,19,18,6, so Sum_{i=1..n} C(i+1,2)^3 = (n+2) * C(n+1,2) * [ a(1,3)/3 + a(2,3)*C(n-1,1)/4 + a(3,3)*C(n-1,2)/5 + a(4,3)*C(n-1,3)/6 + a(5,3)*C(n-1,4)/7 ] = [ (n+2)*(n+1)*n/2 ] * [ 1/3 + (8/4)*C(n-1,1) + (19/5)*C(n-1,2) + (18/6)*C(n-1,3) + (6/7)*C(n-1,4). Cf. A085438 for more details.
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CROSSREFS
| Cf. A000292, A024166, A024166, A085438, A085439, A085440, A085441, A085442, A087107, A000332, A086020, A086021, A086022, A087108, A000389, A086023, A086024, A087109, A000579, A086025, A086026, A087110, A000580, A086027, A086028, A087111, A027555, A086029, A086030.
Sequence in context: A156901 A167400 A165889 * A144946 A157109 A185814
Adjacent sequences: A087124 A087125 A087126 * A087128 A087129 A087130
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KEYWORD
| easy,nonn,tabf
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AUTHOR
| Andre F. Labossiere (boronali(AT)laposte.net), Aug 11 2003
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EXTENSIONS
| Edited by Dean Hickerson (dean.hickerson(AT)yahoo.com), Aug 16 2003
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