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.

A110165 Riordan array (1/sqrt(1-6x+5x^2),(1-3x-sqrt(1-6x+5x^2))/(2x)).

Original entry on oeis.org

1, 3, 1, 11, 6, 1, 45, 30, 9, 1, 195, 144, 58, 12, 1, 873, 685, 330, 95, 15, 1, 3989, 3258, 1770, 630, 141, 18, 1, 18483, 15533, 9198, 3801, 1071, 196, 21, 1, 86515, 74280, 46928, 21672, 7210, 1680, 260, 24, 1, 408105, 356283, 236736, 119154, 44982, 12510, 2484, 333, 27, 1
Offset: 0

Views

Author

Paul Barry, Jul 14 2005

Keywords

Comments

Columns include A026375, A026376 and A026377. Inverse is A110168. Rows sums are A110166. Diagonal sums are A110167.
From Peter Bala, Jan 09 2022: (Start)
This Riordan array has the form ( x*h'(x)/h(x), h(x) ) with h(x) = (1 - 3*x - sqrt(1 - 6*x + 5*x^2))/(2*x) and so belongs to the hitting time subgroup H of the Riordan group (see Peart and Woan).
T(n,k) = [x^(n-k)] f(x)^n with f(x) = 1 + 3*x + x^2. In general the (n,k)-th entry of the hitting time array ( x*h'(x)/h(x), h(x) ) has the form [x^(n-k)] f(x)^n, where f(x) = x/( series reversion of h(x) ). (End)

Examples

			Rows begin
    1;
    3,   1;
   11,   6,   1;
   45,  30,   9,   1;
  195, 144,  58,  12,   1;
  873, 685, 330,  95,  15,   1;
Production array begins:
  3, 1;
  2, 3, 1;
  0, 1, 3, 1;
  0, 0, 1, 3, 1;
  0, 0, 0, 1, 3, 1;
  0, 0, 0, 0, 1, 3, 1;
  0, 0, 0, 0, 0, 1, 3, 1;
  ... - _Philippe Deléham_, Feb 08 2014
		

Programs

  • Maple
    seq(seq( coeff((x^2 + 3*x + 1)^n, x, n-k), k = 0..n ), n = 0..10); # Peter Bala, Jan 09 2022
  • Mathematica
    (* The function RiordanArray is defined in A256893. *)
    RiordanArray[1/Sqrt[1-6#+5#^2]&, (1-3#-Sqrt[1-6#+5#^2])/(2#)&, 10] // Flatten (* Jean-François Alcover, Jul 19 2019 *)

Formula

Number triangle T(n, k) = Sum_{j = 0..n} C(n, j)C(2j, j+k).
T(n,0) = 3*T(n-1,0) + 2*T(n-1,1), T(n,k) = T(n-1,k-1) + 3*T(n-1,k) + T(n-1,k+1) for k > 0, T(0,0) = 1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Jan 24 2014

A080923 First differences of A003946.

Original entry on oeis.org

1, 3, 8, 24, 72, 216, 648, 1944, 5832, 17496, 52488, 157464, 472392, 1417176, 4251528, 12754584, 38263752, 114791256, 344373768, 1033121304, 3099363912, 9298091736, 27894275208, 83682825624, 251048476872, 753145430616
Offset: 0

Views

Author

Paul Barry, Feb 26 2003

Keywords

Comments

Sum of consecutive pairs of elements of A025192.
The alternating sign sequence with g.f. (1-x^2)/(1+3x) gives the diagonal sums of A110168. - Paul Barry, Jul 14 2005
Let M = an infinite lower triangular matrix with the odd integers (1,3,5,...) in every column, with the leftmost column shifted up one row. Then A080923 = lim_{n->inf} M^n. - Gary W. Adamson, Feb 18 2010
a(n+1), n >= 0, with o.g.f. ((1-x^2)/(1-3*x)-1)/x = (3-x)/(1-3*x) provides the coefficients in the formal power series for tan(3*x)/tan(x) = (3-z)/(1-3*z) = Sum_{n>=0} a(n+1)*z^n, with z = tan(x)^2. Convergence holds for 0 <= z < 1/3, i.e., |x| < Pi/6, approximately 0.5235987758. For the numerator and denominator of this o.g.f. see A034867 and A034839, respectively. - Wolfdieter Lang, Jan 18 2013

Crossrefs

Essentially the same as A005051, A026097 and A083583.

Programs

  • Mathematica
    CoefficientList[Series[(1 - x^2) / (1 - 3 x), {x, 0, 20}], x] (* Vincenzo Librandi, Aug 05 2013 *)

Formula

G.f.: (1-x^2)/(1-3*x).
G.f.: 1/(1 - 3*x + x^2 - 3*x^3 + x^4 - 3*x^5 + ...). - Gary W. Adamson, Jan 06 2011
a(n) = 2^3*3^(n-2), n >= 2, a(0) = 1, a(1) = 3. - Wolfdieter Lang, Jan 18 2013
Showing 1-2 of 2 results.