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Triangle read by rows: T(n, k) = (-1)^(n + 1)*L(n) * M(n, k) where M is the inverse of the matrix generated by the triangle A368846 and L(n) is the lcm of the denominators of the terms in the n-th row of M.
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%I #23 Jan 15 2024 10:02:00

%S -1,0,1,0,0,-1,0,0,7,3,0,0,-14,-6,-1,0,0,693,297,55,5,0,0,-30030,

%T -12870,-2431,-260,-15,0,0,4150146,1778634,337480,37310,2625,105,0,0,

%U -21441420,-9189180,-1745458,-194480,-14280,-714,-21

%N Triangle read by rows: T(n, k) = (-1)^(n + 1)*L(n) * M(n, k) where M is the inverse of the matrix generated by the triangle A368846 and L(n) is the lcm of the denominators of the terms in the n-th row of M.

%C As has been observed by T. Curtright, the absolute value of the nonzero terms in row n of the triangle is monotonically decreasing, and the absolute value of each nonzero term T(n, k) is greater than the sum of the absolute value of the terms in the tail of that row.

%C The sum of the n-th row divided by the lcm of the n-th row of A368848 is the Bernoulli number B(2*n).

%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.

%F (Sum_{k=0..n} T(n, k)) / A369135(n) = Bernoulli(2*n).

%F T(n, 2) / T(n, 3) = 7 / 3 for n >= 3.

%e [0] [-1]

%e [1] [0, 1]

%e [2] [0, 0, -1]

%e [3] [0, 0, 7, 3]

%e [4] [0, 0, -14, -6, -1]

%e [5] [0, 0, 693, 297, 55, 5]

%e [6] [0, 0, -30030, -12870, -2431, -260, -15]

%e [7] [0, 0, 4150146, 1778634, 337480, 37310, 2625, 105]

%e [8] [0, 0, -21441420, -9189180, -1745458, -194480, -14280, -714, -21]

%e .

%e For n = 5:

%e (0 + 0 + 693 + 297 + 55 + 5) / 13860 = 5 / 66 = Bernoulli(10).

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

%t Map[# LCM @@ Denominator[#]&, MapIndexed[(-1)^First[#2] Take[#, First[#2]]&, Inverse[PadRight[Table[A368846[n, k], {n, 0, 10}, {k, 0, n}]]]]] (* _Paolo Xausa_, Jan 15 2024 *)

%o (SageMath)

%o M = matrix(ZZ, 32, 32, A368846).inverse()

%o def T(n, k):

%o L = (-1)**(n + 1)*lcm(M[n][k].denominator() for k in range(n + 1))

%o return L * M[n][k]

%o for n in range(9):

%o print([T(n, k) for k in range(n + 1)])

%Y Cf. A368846, A368848, A369135, A369120 (row sums), A369121 (T(n,n)), A369122 (T(n,2)), A000367/A002445 (Bernoulli(2n)).

%K sign,tabl

%O 0,9

%A _Peter Luschny_, Jan 14 2024