A142706 Coefficients of the derivatives of the Eulerian polynomials (with indexing as in A173018).
1, 4, 2, 11, 22, 3, 26, 132, 78, 4, 57, 604, 906, 228, 5, 120, 2382, 7248, 4764, 600, 6, 247, 8586, 46857, 62476, 21465, 1482, 7, 502, 29216, 264702, 624760, 441170, 87648, 3514, 8, 1013, 95680, 1365576, 5241416, 6551770, 2731152, 334880, 8104, 9
Offset: 1
Examples
Triangle T(n, k) starts: { 1}; { 4, 2}; { 11, 22, 3}; { 26, 132, 78, 4}; { 57, 604, 906, 228, 5}; { 120, 2382, 7248, 4764, 600, 6}; { 247, 8586, 46857, 62476, 21465, 1482, 7}; { 502, 29216, 264702, 624760, 441170, 87648, 3514, 8}; {1013, 95680, 1365576, 5241416, 6551770, 2731152, 334880, 8104, 9}.
Programs
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Maple
T := (n, k) -> k * combinat:-eulerian1(n+1, k): for n from 1 to 9 do seq(T(n, k), k = 1..n) od; # Peter Luschny, Feb 07 2023
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Mathematica
T[n_, k_] := Sum[(-1)^j Binomial[n + 1, j](k + 1 - j)^n, {j, 0, k + 1}]; Table[D[Sum[T[n, k]*x^k, {k, 0, n - 1}], x], {n, 1, 10}]; Table[CoefficientList[D[Sum[T[n, k]*x^k, {k, 0, n - 1}], x], x], {n, 1, 10}]; Flatten[%] (* Alternative: *) Needs["Combinatorica`"] Flatten[Table[k*Eulerian[n+1, k], {n, 1, 9}, {k, 1, n}]] (* Peter Luschny, Feb 07 2023 *)
Formula
Let E(n, x) = Sum_{j=0..k} A173018(n, k)*x^k and E'(n, x) = (d/dx) E(x, n). Then T(n, k) = [x^(k-1)] E'(n+1, x).
Extensions
Edited by Peter Luschny, Feb 07 2023