0,2

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.

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.

Rows start (1), (3,0), (7,4,0), (15,30,5,0) etc.

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: A021769 A010600 A098867 this_sequence A065419 A019092 A115870

Adjacent sequences: A062250 A062251 A062252 this_sequence A062254 A062255 A062256

nonn,tabl

Henry Bottomley (se16(AT)btinternet.com), Jun 14 2001