A344910 T(n, k) = denominator([x^k] [z^n] ((1 - i*z)/(1 + i*z))^(i*x)*(1 + z^2)^(-3/4)). Denominators of the coefficients of the symmetric Meixner-Pollaczek polynomials P^(3/4)_{n}(x, Pi/2). Triangle read by rows, T(n, k) for 0 <= k <= n.

1, 1, 1, 4, 1, 1, 1, 6, 1, 3, 32, 1, 6, 1, 3, 1, 80, 1, 3, 1, 15, 128, 1, 720, 1, 18, 1, 45, 1, 2240, 1, 360, 1, 45, 1, 315, 2048, 1, 6720, 1, 720, 1, 45, 1, 315, 1, 322560, 1, 90720, 1, 1080, 1, 945, 1, 2835, 8192, 1, 1612800, 1, 181440, 1, 5400, 1, 1890, 1, 14175

Offset: 0

Author

Peter Luschny, Jul 08 2021

Keywords

• nonn
• tabl
• frac

Examples

			Triangle starts:
  [0] 1;
  [1] 1, 1;
  [2] 4, 1, 1;
  [3] 1, 6, 1, 3;
  [4] 32, 1, 6, 1, 3;
  [5] 1, 80, 1, 3, 1, 15;
  [6] 128, 1, 720, 1, 18, 1, 45;
  [7] 1, 2240, 1, 360, 1, 45, 1, 315;
  [8] 2048, 1, 6720, 1, 720, 1, 45, 1, 315;
  [9] 1, 322560, 1, 90720, 1, 1080, 1, 945, 1, 2835.
		

Links

• R. Koekoek, P. A. Lesky, and R. F. Swarttouw, Hypergeometric Orthogonal Polynomials and Their q-Analogues, Springer, 2010; pp. 213-216.

Crossrefs

Cf. A344909 (numerators).

Cf. A088802 and A123854 (denominator(binomial(1/4, n))) for column 0.

Cf. A049606 (numerator(n!/2^n)) for column n.

Programs

Maple

gf := ((1 - I*z)/(1 + I*z))^(I*x)*(1 + z^2)^(-3/4):
serz := series(gf, z, 22): coeffz := n -> coeff(serz, z, n):
row := n -> seq(denom(coeff(coeffz(n), x, k)), k = 0..n):
seq(row(n), n = 0..10);
# Alternative:
CoeffList := p -> denom(PolynomialTools:-CoefficientList(p, x)):
P := proc(n) option remember; if n = 0 then 1 elif n = 1 then 2*x else
expand((1/n)*(2*x*P(n - 1, x) - (n - 1/2)*P(n - 2, x))) fi end:
ListTools:-Flatten([seq(CoeffList(P(n)), n = 0..10)]);

Mathematica

ForceSimpl[a_] := Collect[Expand[Simplify[FunctionExpand[a]]], x]
f[n_] := I^n Sum[(-1)^k Binomial[-3/4 + I*x, k] Binomial[-3/4 - I*x, n-k], {k, 0, n}] // ForceSimpl;
row[n_] := CoefficientList[f[n], x] // Denominator;
Table[row[n], {n, 0, 10}] // Flatten

Formula

T(n, k) = denominator([x^k] P(n, x), where P(n, x) = i^n*Sum_{k=0..n} (-1)^k* binomial(-3/4 + i*x, k)*binomial(-3/4 - i*x, n - k). The polynomials have the recurrence P(n, x) = (1/n)*(2*x*P(n - 1, x) - (n - 1/2)*P(n - 2, x))), starting with P(0, x) = 1 and P(1, x) = 2*x.