A225076 Triangle read by rows: absolute values of odd-numbered rows of A225356.

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, 1, 177134, 46525293, 1356555432, 9480267666, 19107752148, 9480267666, 1356555432, 46525293, 177134, 1

Offset: 1

Author

Roger L. Bagula, Apr 26 2013

Keywords

• nonn
• tabf

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]

Examples

			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;
  ...
		

Links

• G. C. Greubel, Rows n = 1..50 of the irregular triangle, flattened

Crossrefs

Cf. A002671 (row sums), A034870, A060187, A171692, A225398.

Programs

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}]]

Sage

def A060187(n, k): return sum( (-1)^(k-j)*(2*j-1)^(n-1)*binomial(n, k-j) for j in (1..k) )
def A225076(n,k): return sum( (-1)^(k-j-1)*A060187(2*n, j+1) for j in (0..k-1) )
flatten([[A225076(n, k) for k in (1..2*n-1)] for n in (1..12)]) # G. C. Greubel, Mar 19 2022

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

Extensions

Edited by N. J. A. Sloane, May 06 2013, May 11 2013