A062255 4th level triangle related to Eulerian numbers and binomial transforms (A062254 is third level, 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).

1, 10, 0, 65, 20, 0, 350, 350, 35, 0, 1701, 3696, 1316, 56, 0, 7770, 30660, 24570, 4200, 84, 0, 34105, 220620, 325620, 131020, 12195, 120, 0, 145750, 1447050, 3513345, 2656720, 613140, 33330, 165, 0, 611501, 8901992, 33074448, 41503484, 18444833

Offset

0,2

Comments

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.

Links

Table of n, a(n) for n=0..40.

Formula

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

Example

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

Crossrefs

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: A084033 A340955 A261943 * A285771 A286120 A286774

Adjacent sequences: A062252 A062253 A062254 * A062256 A062257 A062258

Keyword

nonn,tabl

Author

Henry Bottomley, Jun 14 2001

Status

approved