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%I #6 Mar 30 2012 18:51:34
%S 1,10,0,65,20,0,350,350,35,0,1701,3696,1316,56,0,7770,30660,24570,
%T 4200,84,0,34105,220620,325620,131020,12195,120,0,145750,1447050,
%U 3513345,2656720,613140,33330,165,0,611501,8901992,33074448,41503484,18444833
%N 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).
%C 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.
%F A(n, k)=(k+4)*A(n-1, k)+(n-k)*A(n-1, k-1)+A062254(n, k).
%e Rows start (1), (10,0), (65,20,0), (350,350,35,0), etc.
%Y 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).
%K nonn,tabl
%O 0,2
%A _Henry Bottomley_, Jun 14 2001
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