login
A062254
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).
5
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
OFFSET
0,2
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.
FORMULA
A(n, k)=(k+3)*A(n-1, k)+(n-k)*A(n-1, k-1)+A062253(n, k).
EXAMPLE
Rows start (1), (6,0), (25,10,0), (90,120,15,0), etc
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: A272673 A167357 A064381 * A028849 A287470 A138704
KEYWORD
nonn,tabl
AUTHOR
Henry Bottomley, Jun 14 2001
STATUS
approved