OFFSET
0,18
COMMENTS
For n >= 1, T(n, k) equals the number of walks of length k between any two distinct vertices of the complete graph K_(n+1). - Peter Bala, May 30 2024
LINKS
Seiichi Manyama, Antidiagonals n = 0..139, flattened
M. Dukes and C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.
FORMULA
T(n, k) = n^(k-1) - n^(k-2) + n^(k-3) - ... + (-1)^(k-1) = n^(k-1) - T(n, k-1) = n*T(n, k-1) - (-1)^k = (n - 1)*T(n, k-1) + n*T(n, k-2) = round[n^k/(n+1)] for n > 1.
T(n, k) = (-1)^(k+1) * resultant( n*x + 1, (x^k-1)/(x-1) ). - Max Alekseyev, Sep 28 2021
G.f. of row n: x/((1+x) * (1-n*x)). - Seiichi Manyama, Apr 12 2019
E.g.f. of row n: (exp(n*x) - exp(-x))/(n+1). - Stefano Spezia, Feb 20 2024
From Peter Bala, May 31 2024: (Start)
Binomial transform of the m-th row: Sum_{k = 0..n} binomial(n, k)*T(m, k) = (m + 1)^(n-1) for n >= 1.
Let R(m, x) denote the g.f. of the m-th row of the square array. Then R(m_1, x) o R(m_2, x) = R(m_1 + m_2 + m_1*m_2, x), where o denotes the black diamond product of power series as defined by Dukes and White. Cf. A109502.
T(m_1 + m_2 + m_1*m_2, k) = Sum_{i = 0..k} Sum_{j = i..k} binomial(k, i)* binomial(k-i, j-i)*T(m_1, j)*T(m_2, k-i). (End)
EXAMPLE
From Seiichi Manyama, Apr 12 2019: (Start)
Square array begins:
0, 1, -1, 1, -1, 1, -1, 1, ...
0, 1, 0, 1, 0, 1, 0, 1, ...
0, 1, 1, 3, 5, 11, 21, 43, ...
0, 1, 2, 7, 20, 61, 182, 547, ...
0, 1, 3, 13, 51, 205, 819, 3277, ...
0, 1, 4, 21, 104, 521, 2604, 13021, ...
0, 1, 5, 31, 185, 1111, 6665, 39991, ...
0, 1, 6, 43, 300, 2101, 14706, 102943, ... (End)
MAPLE
seq(print(seq((n^k - (-1)^k)/(n+1), k = 0..10)), n = 0..10); # Peter Bala, May 31 2024
MATHEMATICA
T[n_, k_]:=(n^k - (-1)^k)/(n+1); Join[{0}, Table[Reverse[Table[T[n-k, k], {k, 0, n}]], {n, 12}]]//Flatten (* Stefano Spezia, Feb 20 2024 *)
CROSSREFS
KEYWORD
sign,tabl
AUTHOR
Henry Bottomley, Jun 08 2001
STATUS
approved