A342312 T(n, k) = ((2*n + 1)/2)*Sum_{j, k, n} (-1)^(k + j)*(n + j)*binomial(2*n, n - j)* Stirling2(n - k + j, 1 - k + j) with T(0, 0) = 1. Triangle read by row, T(n, k) for 0 <= k <= n.

1, 0, 3, 0, 0, 10, 0, 0, -42, 21, 0, 0, 216, -288, 36, 0, 0, -1320, 3190, -1210, 55, 0, 0, 9360, -34632, 25584, -4056, 78, 0, 0, -75600, 389340, -462000, 152460, -11970, 105, 0, 0, 685440, -4621824, 7907040, -4368320, 762960, -32640, 136

Offset: 0

Author

Peter Luschny, Mar 08 2021

Keywords

• sign
• tabl
• frac

Comments

The triangle can be seen as representing the numerators of a sequence of rational polynomials. Let p_{n}(x) = Sum_{k=0..n} (T(n, k)/A342313(n, k))*x^k. Then p_{n}(1) = B_{n}(1), where B_{n}(x) are the Bernoulli polynomials.

Examples

			The triangle starts:
  [0] 1
  [1] 0, 3
  [2] 0, 0,    10
  [3] 0, 0,   -42,     21
  [4] 0, 0,   216,   -288,    36
  [5] 0, 0, -1320,   3190, -1210,    55
  [6] 0, 0,  9360, -34632, 25584, -4056, 78
The first few polynomials are P(n, k) = T(n, k) / A342313(n, k):
  1;
  0, 1/2;
  0,  0,  1/6;
  0,  0, -1/10,   1/10;
  0,  0,  3/35,  -4/21,   1/14;
  0,  0, -2/21,  29/84, -11/36,    1/18;
  0,  0, 10/77, -37/55, 164/165, -26/55, 1/22;
		

Links

• Levin Luschny, Illustrating the polynomials.

Crossrefs

Cf. A014105 (main diagonal), A342313 (denominators), A340556.

Programs

Maple

T := (n, k) -> `if`(n = 0 and k = 0, 1, (n+1/2)*
add((-1)^j*(n+k+j)*binomial(2*n, n-k-j)*Stirling2(n + j, j + 1) , j= 0..n-k)):
seq(print(seq(T(n, k), k=0..n)), n=0..6);

Mathematica

T[0, 0] := 1; T[n_, k_] := ((2n + 1)/2) Sum[(-1)^(k+j)(n+j) Binomial[2n, n-j] StirlingS2[n-k+j, 1-k+j], {j, k, n}];
Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten

Formula

T(n, k) = numerator([x^k] p(n, x)) for n >= 2, where p(n, x) = (1/2)*Sum_{k=0..n-1} (-1)^k*x^(n-k)*E2(n - 1, k + 1) / binomial(2*n - 1, k + 1) and E2(n,k) denotes the second-order Eulerian numbers A340556.

Another representation of the polynomials for n >= 2 is p(n, x) = (1/2)*Sum_{k=0..n} x^k*Sum_{j=k..n} ((-1)^(j + k)*((n - k + 1)!*(n + k - 2)!)/((j + n - 1)!*(n - j)!))*Stirling2(n - k + j, j - k + 1).