0,2

Binomial transform 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)^4.

A(n, k)=(k+4)*A(n-1, k)+(n-k)*A(n-1, k-1)+A062254(n, k).

Rows start (1), (10,0), (65,20,0), (350,350,35,0), etc.

First column is A000453. Diagonals include A000007, and all but the start of A000292. Row sums are A000454. 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: A087474 A051906 A084033 this_sequence A092916 A089831 A070190

Adjacent sequences: A062252 A062253 A062254 this_sequence A062256 A062257 A062258

nonn,tabl

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