login
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.
3
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
0,3
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.
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).
EXAMPLE
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;
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
CROSSREFS
Cf. A014105 (main diagonal), A342313 (denominators), A340556.
Sequence in context: A346811 A244127 A363407 * A123474 A370064 A363033
KEYWORD
sign,tabl,frac
AUTHOR
Peter Luschny, Mar 08 2021
STATUS
approved