cp's OEIS Frontend

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).

%I A062255 #10 Jan 27 2025 10:00:46
%S A062255 1,10,0,65,20,0,350,350,35,0,1701,3696,1316,56,0,7770,30660,24570,
%T A062255 4200,84,0,34105,220620,325620,131020,12195,120,0,145750,1447050,
%U A062255 3513345,2656720,613140,33330,165,0,611501,8901992,33074448,41503484,18444833,2634192,87406,220,0
%N 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).
%C A062255 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 A062255 A(n, k) = (k+4)*A(n-1, k)+(n-k)*A(n-1, k-1) + A062254(n, k).
%e A062255 Rows start:
%e A062255  (1),
%e A062255  (10,0),
%e A062255  (65,20,0),
%e A062255  (350,350,35,0), etc.
%o A062255 (PARI) E(n, k) = if ((n<0) || (k<0), 0, if ((n==0) && (k==0), 1, (k+1)*E(n-1, k)+(n-k)*E(n-1, k-1)));
%o A062255 A2(n, k) = if ((n<0) || (k<0), 0, (k+2)*A2(n-1, k)+(n-k)*A2(n-1, k-1)+E(n, k));
%o A062255 A3(n, k) = if ((n<0) || (k<0), 0, (k+3)*A3(n-1, k)+(n-k)*A3(n-1, k-1) + A2(n, k));
%o A062255 A4(n, k) = if ((n<0) || (k<0), 0, (k+4)*A4(n-1, k)+(n-k)*A4(n-1, k-1)+ A3(n, k));
%o A062255 row4(n) = vector(n+1, k, A4(n,k-1)); \\ _Michel Marcus_, Jan 27 2025
%Y A062255 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 A062255 nonn,tabl
%O A062255 0,2
%A A062255 _Henry Bottomley_, Jun 14 2001
%E A062255 More terms from _Michel Marcus_, Jan 27 2025