cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A114193 Riordan array (1/(1+2xc(-2x)),xc(-2x)/(1+2xc(-2x))), c(x) the g.f. of A000108.

Original entry on oeis.org

1, -2, 1, 8, -6, 1, -40, 36, -10, 1, 224, -224, 80, -14, 1, -1344, 1440, -600, 140, -18, 1, 8448, -9504, 4400, -1232, 216, -22, 1, -54912, 64064, -32032, 10192, -2184, 308, -26, 1, 366080, -439296, 232960, -81536, 20160, -3520, 416, -30, 1, -2489344, 3055104, -1697280, 639744, -176256, 35904, -5304, 540, -34, 1
Offset: 0

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Author

Paul Barry, Nov 16 2005

Keywords

Comments

Row sums are A114191. Diagonal sums are A114194. Inverse of A114192.
Triangle T(n,k), read by rows, given by (-2, -2, -2, -2, -2, -2, -2, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Jan 26 2014

Examples

			Triangle begins
      1;
     -2,    1;
      8,   -6,    1;
    -40,   36,  -10,   1;
    224, -224,   80, -14,   1;
  -1344, 1440, -600, 140, -18, 1;
		

Crossrefs

Programs

  • Mathematica
    c[x_] := (1 - Sqrt[1 - 4 x])/(2 x);
    (* The function RiordanArray is defined in A256893. *)
    RiordanArray[1/(1 + 2 # c[-2 #])&, # c[-2 #]/(1 + 2 # c[-2 #])&, 10] // Flatten (* Jean-François Alcover, Jul 16 2019 *)

Formula

Riordan array ((sqrt(1+8x)-1)/(4x), (sqrt(1+8x)-1)^2/(16x)).
T(n, k) = (-2)^(n-k)*A039599(n, k) = (-2)^(n-k)*C(2*n, n-k)*(2*k+1)/(n+k+1). - Philippe Deléham, Nov 17 2005

A329918 Coefficients of orthogonal polynomials related to the Jacobsthal numbers A152046, triangle read by rows, T(n, k) for 0 <= k <= n.

Original entry on oeis.org

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

Views

Author

Peter Luschny, Nov 28 2019

Keywords

Examples

			Triangle starts:
  [0] 1;
  [1] 0,  1;
  [2] 0,  0,  1;
  [3] 0,  2,  0,  1;
  [4] 0,  0,  4,  0,  1;
  [5] 0,  4,  0,  6,  0,  1;
  [6] 0,  0, 12,  0,  8,  0,  1;
  [7] 0,  8,  0, 24,  0, 10,  0,  1;
  [8] 0,  0, 32,  0, 40,  0, 12,  0, 1;
  [9] 0, 16,  0, 80,  0, 60,  0, 14, 0, 1;
The first few polynomials:
  p(0,x) = 1;
  p(1,x) = x;
  p(2,x) = x^2;
  p(3,x) = 2*x + x^3;
  p(4,x) = 4*x^2 + x^4;
  p(5,x) = 4*x + 6*x^3 + x^5;
  p(6,x) = 12*x^2 + 8*x^4 + x^6;
		

Crossrefs

Row sums are A001045 starting with 1, which is A152046. These are in signed form also the alternating row sums. Diagonal sums are aerated A133494.

Programs

  • Julia
    using Nemo # Returns row n.
    function A329918(row)
        R, x = PolynomialRing(ZZ, "x")
        function p(n)
            n < 3 && return x^n
            x*p(n-1) + 2*p(n-2)
        end
        p = p(row)
        [coeff(p, k) for k in 0:row]
    end
    for row in 0:9 println(A329918(row)) end # prints triangle
  • Maple
    T := (n, k) -> `if`((n+k)::odd, 0, 2^((n-k)/2)*binomial((n+k)/2-1, (n-k)/2)):
    seq(seq(T(n, k), k=0..n), n=0..11);

Formula

p(n) = x*p(n-1) + 2*p(n-2) for n >= 3; p(0) = 1, p(1) = x, p(2) = x^2.
T(n, k) = [x^k] p(n).
T(n, k) = 2^((n-k)/2)*binomial((n+k)/2-1, (n-k)/2) if n+k is even otherwise 0.
Showing 1-2 of 2 results.