OFFSET
1,3
COMMENTS
An equivalent definition: take the polynomials corresponding to rows 2, 4, 6, 8, ... of A060187, divide by x+1, and extract the coefficients. [Corrected by Petros Hadjicostas, Apr 17 2020]
LINKS
G. C. Greubel, Rows n = 1..50 of the irregular triangle, flattened
FORMULA
Triangle read by rows: row n gives coefficients in the expansion of the polynomial ((x - 1)^(2*n)/(x + 1)) * Sum_{k >= 0} (2*k + 1)^(2*n-1)*x^k. The infinite sum simplifies to a polynomial.
Sum_{m=0..2*n-2} T(n,m)*t^m = 2^(2*n-1) * (1-t)^(2*n) * LerchPhi(t, 1-2*n, 1/2)/(1 + t).
Sum_{k=1..n} T(n, k) = A002671(n-1).
T(n,m) = Sum_{k=0..m-1} (-1)^(m-k-1)*A060187(2*n,k+1) for n >= 1 and 1 <= m <= 2*n-1. - Petros Hadjicostas, Apr 17 2020
EXAMPLE
Triangle T(n,m) (for n >= 1 and 0 <= m <= 2*n - 2) begins as follows:
1;
1, 22, 1;
1, 236, 1446, 236, 1;
1, 2178, 58479, 201244, 58479, 2178, 1;
1, 19672, 1736668, 19971304, 49441990, 19971304, 1736668, 19672, 1;
...
MATHEMATICA
(* Power series via an infinite sum *)
p[x_, n_] = (x-1)^(2*n)*Sum[(2*k+1)^(2*n-1)*x^k, {k, 0, Infinity}];
Table[CoefficientList[p[x, n]/(1+x), x], {n, 1, 10}]//Flatten
(* First alternative method: recurrence *)
t[n_, k_, m_]:= t[n, k, m]= If[k==1 || k==n, 1, (m*n-m*k+1)*t[n-1, k-1, m] + (m*k - (m-1))*t[n-1, k, m]];
T[n_, k_]:= T[n, k]= t[n+1, k+1, 2]; (* t(n, k, 2) = A060187 *)
Flatten[Table[CoefficientList[Sum[T[n, k]*x^k, {k, 0, n}]/(x+1), x], {n, 14, 2}]]
(* Second alternative method: polynomial expansion *)
p[t_] = Exp[t]*x/(-Exp[2*t] + x);
Flatten[Table[CoefficientList[(n!*(-1+x)^(n+1)/(x*(x+1)))*SeriesCoefficient[ Series[p[t], {t, 0, 30}], n], x], {n, 1, 13, 2}]]
PROG
(Sage)
def A060187(n, k): return sum( (-1)^(k-j)*(2*j-1)^(n-1)*binomial(n, k-j) for j in (1..k) )
flatten([[A225076(n, k) for k in (1..2*n-1)] for n in (1..12)]) # G. C. Greubel, Mar 19 2022
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Roger L. Bagula, Apr 26 2013
EXTENSIONS
Edited by N. J. A. Sloane, May 06 2013, May 11 2013
STATUS
approved