OFFSET
0,9
COMMENTS
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.
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).
LINKS
Thomas Curtright, Scale Invariant Scattering and the Bernoulli Numbers, arXiv:2401.00586 [math-ph], Jan 2024.
FORMULA
(Sum_{k=0..n} T(n, k)) / A369135(n) = Bernoulli(2*n).
T(n, 2) / T(n, 3) = 7 / 3 for n >= 3.
EXAMPLE
[0] [-1]
[1] [0, 1]
[2] [0, 0, -1]
[3] [0, 0, 7, 3]
[4] [0, 0, -14, -6, -1]
[5] [0, 0, 693, 297, 55, 5]
[6] [0, 0, -30030, -12870, -2431, -260, -15]
[7] [0, 0, 4150146, 1778634, 337480, 37310, 2625, 105]
[8] [0, 0, -21441420, -9189180, -1745458, -194480, -14280, -714, -21]
.
For n = 5:
(0 + 0 + 693 + 297 + 55 + 5) / 13860 = 5 / 66 = Bernoulli(10).
MATHEMATICA
A368846[n_, k_] := If[k == 0, Boole[n == 0], (-1)^(n + k) 2 Binomial[2 k - 1, n] Binomial[2 n + 1, 2 k]];
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 *)
PROG
(SageMath)
M = matrix(ZZ, 32, 32, A368846).inverse()
def T(n, k):
L = (-1)**(n + 1)*lcm(M[n][k].denominator() for k in range(n + 1))
return L * M[n][k]
for n in range(9):
print([T(n, k) for k in range(n + 1)])
CROSSREFS
KEYWORD
sign,tabl
AUTHOR
Peter Luschny, Jan 14 2024
STATUS
approved