cp's OEIS Frontend

A062254 3rd level triangle related to Eulerian numbers and binomial transforms (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 A062254 #13 Jan 27 2025 07:30:21
%S A062254 1,6,0,25,10,0,90,120,15,0,301,896,406,21,0,966,5376,5586,1176,28,0,
%T A062254 3025,28470,55560,27910,3123,36,0,9330,139320,456525,437100,122520,
%U A062254 7860,45,0,28501,646492,3312078,5339719,2912833,494802,19096,55,0
%N A062254 3rd level triangle related to Eulerian numbers and binomial transforms (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 A062254 Binomial transform 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)^3.
%F A062254 A(n, k) = (k+3)*A(n-1, k) + (n-k)*A(n-1, k-1) + A062253(n, k).
%e A062254 Rows start:
%e A062254  (1),
%e A062254  (6,0),
%e A062254  (25,10,0),
%e A062254  (90,120,15,0),
%e A062254  ...
%o A062254 (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 A062254 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 A062254 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 A062254 row3(n) = vector(n+1, k, A3(n,k-1)); \\ _Michel Marcus_, Jan 27 2025
%Y A062254 First column is A000392. Diagonals include A000007 and all but the start of A000217. Row sums are A000399.
%Y A062254 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, (cont.)
%Y A062254 (cont.) 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 A062254 nonn,tabl
%O A062254 0,2
%A A062254 _Henry Bottomley_, Jun 14 2001