A136688 Triangle of polynomials F(x,n) = x*F(x,n-1) + 2*F(x,n-2).

1, 0, 1, 2, 0, 1, 0, 4, 0, 1, 4, 0, 6, 0, 1, 0, 12, 0, 8, 0, 1, 8, 0, 24, 0, 10, 0, 1, 0, 32, 0, 40, 0, 12, 0, 1, 16, 0, 80, 0, 60, 0, 14, 0, 1, 0, 80, 0, 160, 0, 84, 0, 16, 0, 1, 32, 0, 240, 0, 280, 0, 112, 0, 18, 0, 1, 0, 192, 0, 560, 0, 448, 0, 144, 0, 20, 0, 1

Offset

1,4

Comments

Riordan array (1/(1-2*x^2), x/(1-2*x^2)). - Paul Barry, Jun 18 2008

Antidiagonal sums are 1,0,3,0,9,... with g.f. 1/(1-3*x^2). - Paul Barry, Jun 18 2008

Links

G. C. Greubel, Rows n = 1..100 of triangle, flattened (terms 1..500 from Nathaniel Johnston)

J. Cigler, q-Fibonacci polynomials, Fibonacci Quarterly 41 (2003) 31-40.

Formula

F(x,n) = x*F(x,n-1) + s*F(x,n-2), where F(x,0)=0, F(x,1)=1 and s=2.

F(x,n) = Sum_{j=0..floor((n-1)/2)} binomial(n-j-1, j)*x^(n-2*j-1)*2^j, for n >= 1. See the Mma program by G. C. Greubel. - Wolfdieter Lang, Feb 10 2023

From Andrew Howroyd, Feb 11 2023: (Start)

T(n,k) = 2^((n-k-1)/2)*binomial((n+k-1)/2, (n-k-1)/2) for k+1 == n (mod 2).

G.f.: x/(1 - y*x - 2*x^2). (End)

Example

Triangle begins:
   1;
   0,   1;
   2,   0,   1;
   0,   4,   0,   1;
   4,   0,   6,   0,   1;
   0,  12,   0,   8,   0,   1;
   8,   0,  24,   0,  10,   0,   1;
   0,  32,   0,  40,   0,  12,   0,   1;
  16,   0,  80,   0,  60,   0,  14,   0,   1;
   0,  80,   0, 160,   0,  84,   0,  16,   0,   1;
  32,   0, 240,   0, 280,   0, 112,   0,  18,   0,   1;
  ...

Maple

A136688 := proc(n) option remember: if(n<=1)then return n: else return x*A136688(n-1)+2*A136688(n-2): fi: end:
seq(seq(coeff(A136688(n), x, m), m=0..n-1), n=1..10); # Nathaniel Johnston, Apr 27 2011

Mathematica

s = 2; F[x_, n_]:= F[x, n]= If[n<2, n, x*F[x, n-1] + s*F[x, n-2]]; Table[CoefficientList[F[x, n], x], {n, 12}]//Flatten
F[n_, x_, s_, q_]:= Sum[QBinomial[n-j-1, j, q]*q^Binomial[j+1, 2]*x^(n-2*j-1) *s^j, {j, 0, Floor[(n-1)/2]}]; Table[CoefficientList[F[n, x, 2, 1], x], {n, 1, 10}]//Flatten (* G. C. Greubel, Dec 16 2019 *)

Prog

(Sage)
def f(n, x, s, q): return sum( q_binomial(n-j-1, j, q)*q^binomial(j+1, 2)*x^(n-2*j-1)*s^j for j in (0..floor((n-1)/2)))
def A136688_list(prec):
    P.<x> = PowerSeriesRing(ZZ, prec)
    return P( f(n, x, 2, 1) ).list()
[A136688_list(n) for n in (1..10)] # G. C. Greubel, Dec 16 2019
(PARI) T(n, k)=if((n-k)%2==0, 0, 2^((n-k-1)/2)*binomial((n+k-1)/2, (n-k-1)/2)) \\ Andrew Howroyd, Feb 11 2023

Crossrefs

Row sums give A001045.

Cf. A136689, A136705.

Sequence in context: A053389 A354667 A202328 * A131321 A111959 A110109

Adjacent sequences: A136685 A136686 A136687 * A136689 A136690 A136691

Keyword

nonn,easy,tabl

Author

Roger L. Bagula, Apr 06 2008

Status

approved