login
Triangle read by rows: T(n, k) = (-1)^(n + k)*2*binomial(2*k - 1, n)* binomial(2*n + 1, 2*k) for k > 0, and k^n for k = 0.
6

%I #23 Jan 14 2024 09:11:41

%S 1,0,6,0,0,30,0,0,-70,140,0,0,0,-840,630,0,0,0,924,-6930,2772,0,0,0,0,

%T 18018,-48048,12012,0,0,0,0,-12870,216216,-300300,51480,0,0,0,0,0,

%U -350064,2042040,-1750320,218790,0,0,0,0,0,184756,-5542680,16628040,-9699690,923780

%N Triangle read by rows: T(n, k) = (-1)^(n + k)*2*binomial(2*k - 1, n)* binomial(2*n + 1, 2*k) for k > 0, and k^n for k = 0.

%C The row sums of the inverse triangle (A368847/A368848) are the unsigned Bernoulli numbers |B(2n)|. To get the signed Bernoulli numbers B(2n), one only needs to change the sign factor in the definition from (-1)^(n + k) to (-1)^(n + 1).

%C Conjecture: |Sum_{j=0..k} T(k + j, k)| = A229580(k + 1) for k >= 0.

%H Paolo Xausa, <a href="/A368846/b368846.txt">Table of n, a(n) for n = 0..11475</a> (rows 0..150 of the triangle, flattened).

%H Thomas Curtright, <a href="https://doi.org/10.48550/arXiv.2401.00586">Scale Invariant Scattering and the Bernoulli Numbers</a>, arXiv:2401.00586 [math-ph], Jan 2024.

%e [0] [1]

%e [1] [0, 6]

%e [2] [0, 0, 30]

%e [3] [0, 0, -70, 140]

%e [4] [0, 0, 0, -840, 630]

%e [5] [0, 0, 0, 924, -6930, 2772]

%e [6] [0, 0, 0, 0, 18018, -48048, 12012]

%e [7] [0, 0, 0, 0, -12870, 216216, -300300, 51480]

%e [8] [0, 0, 0, 0, 0, -350064, 2042040, -1750320, 218790]

%t A368846[n_,k_] := If[k==0, Boole[n==0], (-1)^(n+k) 2 Binomial[2k-1, n] Binomial[2n+1, 2k]];

%t Table[A368846[n, k], {n,0,10}, {k,0,n}] (* _Paolo Xausa_, Jan 08 2024 *)

%o (SageMath)

%o def A368846(n, k):

%o if k == 0: return k^n

%o if k > n: return 0

%o return (-1)^(n + k)*2*binomial(2*k - 1, n)*binomial(2*n + 1, 2*k)

%o for n in range(10): print([A368846(n, k) for k in range(n+1)])

%Y Cf. A368847/A368848 (inverse), A369134, A369135, A002457 (main diagonal), A000367/A002445 (Bernoulli(2n)), A229580.

%K sign,tabl

%O 0,3

%A _Peter Luschny_, Jan 07 2024