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Search: id:A089463
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| A089463 |
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Triangle, read by rows, of coefficients for the third iteration of the hyperbinomial transform. |
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+0
5
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| 1, 3, 1, 15, 6, 1, 108, 45, 9, 1, 1029, 432, 90, 12, 1, 12288, 5145, 1080, 150, 15, 1, 177147, 73728, 15435, 2160, 225, 18, 1, 3000000, 1240029, 258048, 36015, 3780, 315, 21, 1, 58461513, 24000000, 4960116, 688128, 72030, 6048, 420, 24, 1, 1289945088
(list; table; graph; listen)
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OFFSET
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0,2
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COMMENT
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Equals the matrix cube of A088956 when treated as a lower triangular matrix. The 3rd hyperbinomial transform of a sequence {b} is defined to be the sequence {d} given by d(n) = sum(k=0..n, T(n,k)*b(k)), where T(n,k) = 3*(n-k+3)^(n-k-1)*C(n,k). Given a table in which the n-th row is the n-th binomial transform of the first row, then the 3rd hyperbinomial transform of any diagonal results in the 3rd diagonal lower in the table.
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FORMULA
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T(n, k) = 3*(n-k+3)^(n-k-1)*C(n, k). E.g.f.: exp(x*y)*(-LambertW(-y)/y)^3. Note: (-LambertW(-y)/y)^3 = sum(n>=0, 3*(n+3)^(n-1)*y^n/n!).
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EXAMPLE
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Rows begin:
{1},
{3,1},
{15,6,1},
{108,45,9,1},
{1029,432,90,12,1},
{12288,5145,1080,150,15,1},
{177147,73728,15435,2160,225,18,1},
{3000000,1240029,258048,36015,3780,315,21,1},..
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CROSSREFS
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Cf. A089464(row sums), A089465(diagonal), A089460, A088956.
Sequence in context: A092589 A048966 A104990 this_sequence A136231 A113389 A038553
Adjacent sequences: A089460 A089461 A089462 this_sequence A089464 A089465 A089466
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KEYWORD
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nonn,tabl
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AUTHOR
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Paul D. Hanna (pauldhanna(AT)juno.com), Nov 05 2003
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