cp's OEIS Frontend

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.

%I A342312 #20 Apr 12 2025 09:14:57
%S A342312 1,0,3,0,0,10,0,0,-42,21,0,0,216,-288,36,0,0,-1320,3190,-1210,55,0,0,
%T A342312 9360,-34632,25584,-4056,78,0,0,-75600,389340,-462000,152460,-11970,
%U A342312 105,0,0,685440,-4621824,7907040,-4368320,762960,-32640,136
%N 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.
%C A342312 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.
%H A342312 Levin Luschny, <a href="/A342312/a342312.png">Illustrating the polynomials</a>.
%F A342312 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.
%F A342312 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).
%e A342312 The triangle starts:
%e A342312   [0] 1
%e A342312   [1] 0, 3
%e A342312   [2] 0, 0,    10
%e A342312   [3] 0, 0,   -42,     21
%e A342312   [4] 0, 0,   216,   -288,    36
%e A342312   [5] 0, 0, -1320,   3190, -1210,    55
%e A342312   [6] 0, 0,  9360, -34632, 25584, -4056, 78
%e A342312 The first few polynomials are P(n, k) = T(n, k) / A342313(n, k):
%e A342312   1;
%e A342312   0, 1/2;
%e A342312   0,  0,  1/6;
%e A342312   0,  0, -1/10,   1/10;
%e A342312   0,  0,  3/35,  -4/21,   1/14;
%e A342312   0,  0, -2/21,  29/84, -11/36,    1/18;
%e A342312   0,  0, 10/77, -37/55, 164/165, -26/55, 1/22;
%p A342312 T := (n, k) -> `if`(n = 0 and k = 0, 1, (n+1/2)*
%p A342312 add((-1)^j*(n+k+j)*binomial(2*n, n-k-j)*Stirling2(n + j, j + 1) , j= 0..n-k)):
%p A342312 seq(print(seq(T(n, k), k=0..n)), n=0..6);
%t A342312 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}];
%t A342312 Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten
%Y A342312 Cf. A014105 (main diagonal), A342313 (denominators), A340556.
%K A342312 sign,tabl,frac
%O A342312 0,3
%A A342312 _Peter Luschny_, Mar 08 2021