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A062254
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3rd level triangle related to Eulerian numbers and binomial transforms (A062253 is second level, triangle of Eulerian numbers is first level and triangle with Z(0,0)=1 and Z(n,k)=0 otherwise is 0th level).
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3
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1, 6, 0, 25, 10, 0, 90, 120, 15, 0, 301, 896, 406, 21, 0, 966, 5376, 5586, 1176, 28, 0, 3025, 28470, 55560, 27910, 3123, 36, 0, 9330, 139320, 456525, 437100, 122520, 7860, 45, 0, 28501, 646492, 3312078, 5339719, 2912833, 494802, 19096, 55, 0
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| Binomial transform of n^3*k^n is ((kn)^3 + 3(kn)^2 + (1 - k)(kn))*(k + 1)^(n - 3); of n^4*k^n is ((kn)^4 + 6(kn)^3 + (7 - 4k)(kn)^2 + (1 - 4k + k^2)(kn))*(k + 1)^(n - 4); of n^5*k^n is ((kn)^5 + 10(kn)^4 + (25 - 10k)(kn)^3 + (15 - 30k + 5k^2)(kn)^2 + (1 - 11k + 11k^2 - k^3)(kn))*(k + 1)^(n - 5); of n^6*k^n is ((kn)^6 + 15(kn)^5 + (65 - 20k)(kn)^4 + (90 - 120k + 15k^2)(kn)^3 + (31 - 146k + 91k^2 - 6k^3)(kn)^2 + (1 - 26k + 66k^2 - 26k^3 + k^4)(kn))*(k + 1)^(n - 6). This sequence gives the (unsigned) polynomial coefficients of (kn)^3.
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FORMULA
| A(n, k)=(k+3)*A(n-1, k)+(n-k)*A(n-1, k-1)+A062253(n, k).
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EXAMPLE
| Rows start (1), (6,0), (25,10,0), (90,120,15,0), etc
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CROSSREFS
| First column is A000392. Diagonals include A000007 and all but the start of A000217. Row sums are A000399.
Taking all the levels together to create a pyramid, one face would be A010054 as a triangle with a parallel face which is Pascal's triangle (A007318) with two columns removed, another face would be a triangle of Stirling numbers of the second kind (A008277) and a third face would be A000007 as a triangle, (cont.)
(cont.) with a triangle of Eulerian numbers (A008292), A062253, A062254 and A062255 as faces parallel to it. The row sums of this last group would provide a triangle of unsigned Stirling numbers of the first kind (A008275).
Sequence in context: A082731 A167357 A064381 * A028849 A138704 A005396
Adjacent sequences: A062251 A062252 A062253 * A062255 A062256 A062257
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KEYWORD
| nonn,tabl
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AUTHOR
| Henry Bottomley (se16(AT)btinternet.com), Jun 14 2001
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