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 A089463 Triangle, read by rows, of coefficients for the third iteration of the hyperbinomial transform. 5
 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; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS 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. LINKS G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened FORMULA 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!). EXAMPLE 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},.. MATHEMATICA Flatten[Table[3(n-k+3)^(n-k-1) Binomial[n, k], {n, 0, 10}, {k, 0, n}]] (* Harvey P. Dale, Jun 26 2013 *) PROG (PARI) for(n=0, 10, for(k=0, n, print1(3*(n-k+3)^(n-k-1)*binomial(n, k), ", "))) \\ G. C. Greubel, Nov 17 2017 CROSSREFS Cf. A089464(row sums), A089465(diagonal), A089460, A088956. Sequence in context: A048966 A297704 A104990 * A136231 A113389 A038553 Adjacent sequences:  A089460 A089461 A089462 * A089464 A089465 A089466 KEYWORD nonn,tabl AUTHOR Paul D. Hanna, Nov 05 2003 STATUS approved

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Last modified September 21 19:57 EDT 2020. Contains 337273 sequences. (Running on oeis4.)