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A062253
2nd level triangle related to Eulerian numbers and binomial transforms (triangle of Eulerian numbers is first level and triangle with Z(0,0)=1 and Z(n,k)=0 otherwise is 0th level).
6
1, 3, 0, 7, 4, 0, 15, 30, 5, 0, 31, 146, 91, 6, 0, 63, 588, 868, 238, 7, 0, 127, 2136, 6126, 4096, 575, 8, 0, 255, 7290, 36375, 47400, 16929, 1326, 9, 0, 511, 23902, 193533, 434494, 306793, 64362, 2971, 10, 0, 1023, 76296, 956054, 3421902, 4169418
OFFSET
0,2
COMMENTS
Binomial transform of n^2*k^n is ((kn)^2 + kn)*(k + 1)^(n - 2); 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)^2.
FORMULA
A(n, k)=(k+2)*A(n-1, k)+(n-k)*A(n-1, k-1)+E(n, k) where E(n, k)=(k+1)*E(n-1, k)+(n-k)*E(n-1, k-1) and E(0, 0)=1 is a triangle of Eulerian numbers, essentially A008292.
EXAMPLE
Rows start (1), (3,0), (7,4,0), (15,30,5,0) etc.
CROSSREFS
First column is A000225. Diagonals include A000007, A009056. Row sums are A000254. 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, 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: A181163 A363392 A098867 * A290844 A369094 A322018
KEYWORD
nonn,tabl
AUTHOR
Henry Bottomley, Jun 14 2001
STATUS
approved