A236830 Riordan array (1/(1-x*C(x)^3), x*C(x)), C(x) the g.f. of A000108.

1, 1, 1, 4, 2, 1, 16, 7, 3, 1, 65, 27, 11, 4, 1, 267, 108, 43, 16, 5, 1, 1105, 440, 173, 65, 22, 6, 1, 4597, 1812, 707, 267, 94, 29, 7, 1, 19196, 7514, 2917, 1105, 398, 131, 37, 8, 1, 80380, 31307, 12111, 4597, 1680, 575, 177, 46, 9, 1, 337284, 130883, 50503, 19196, 7085, 2488, 808, 233, 56, 10, 1

Offset: 0

Author

Philippe Deléham, Feb 01 2014

Keywords

• nonn
• tabl

Comments

T(n+3,n) = A011826(n+5).

Examples

			Triangle begins:
      1;
      1,    1;
      4,    2,    1;
     16,    7,    3,    1;
     65,   27,   11,    4,   1;
    267,  108,   43,   16,   5,   1;
   1105,  440,  173,   65,  22,   6,  1;
   4597, 1812,  707,  267,  94,  29,  7, 1;
  19196, 7514, 2917, 1105, 398, 131, 37, 8, 1;
Production matrix is:
   1  1
   3  1   1
   6  1   1   1
  10  1   1   1   1
  15  1   1   1   1   1
  21  1   1   1   1   1   1
  28  1   1   1   1   1   1   1
  36  1   1   1   1   1   1   1   1
  45  1   1   1   1   1   1   1   1   1
  55  1   1   1   1   1   1   1   1   1   1
  66  1   1   1   1   1   1   1   1   1   1   1
  78  1   1   1   1   1   1   1   1   1   1   1   1
  91  1   1   1   1   1   1   1   1   1   1   1   1   1
  ...
		

Links

• G. C. Greubel, Rows n= 0..100 of triangle, flattened

Crossrefs

Cf. (columns): A165201, A026726, A026671, A026674, A026672, A026675, A026673, A026843, A026842 or A026846 or A026849, A026844, A026841 or A026848.

Programs

GAP

Flat(List([0..12], n-> List([0..n], k-> Sum([0..n-k], j-> Binomial(2*n-k-j-2, n-k-j)*Fibonacci(2*j-1) )))); # G. C. Greubel, Jul 18 2019

Magma

[(&+[Binomial(2*n-k-j-2, n-k-j)*Fibonacci(2*j-1): j in [0..n-k]]): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 18 2019

Maple

A236830 := (n,k) -> add(combinat:-fibonacci(2*i-1)*binomial(2*n-2-k-i,n-k-i), i = 0..n-k): seq(seq(A236830(n, k), k = 0..n), n = 0..10); # Peter Bala, Feb 18 2018

Mathematica

(* The function RiordanArray is defined in A256893. *)
c[x_] := (1 - Sqrt[1 - 4 x])/(2 x);
RiordanArray[1/(1 - # c[#]^3)&, # c[#]&, 11] // Flatten (* Jean-François Alcover, Jul 16 2019 *)
Table[Sum[Binomial[2*n-k-j-2, n-k-j]*Fibonacci[2*j-1], {j,0,n-k}], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Jul 18 2019 *)

PARI

T(n,k) = sum(j=0,n-k, binomial(2*n-k-j-2, n-k-j)*fibonacci(2*j -1));
for(n=0,12, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Jul 18 2019

Sage

[[sum( binomial(2*n-k-j-2, n-k-j)*fibonacci(2*j -1) for j in (0..n-k) ) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jul 18 2019

Formula

Sum_{k=0..n} T(n,k) = A026726(n).

G.f.: 1/((x^2*C(x)^4-x*C(x))*y-x*C(x)^3+1), where C(x) the g.f. of A000108. - Vladimir Kruchinin, Apr 22 2015

From Peter Bala, Feb 18 2018: (Start)

T(n,k) = Sum_{i = 0..n-k} Fibonacci(2*i-1)*binomial(2*n-2-k-i,n-k-i).

The n-th row polynomial of row reverse triangle is the n-th degree Taylor polynomial of the rational function (1 - 3*x + 2*x^2)/(1 - 3*x + x^2) * 1/(1 - x)^n about 0. For example, for n = 4, (1 - 3*x + 2*x^2)/(1 - 3*x + x^2) * 1/(1 - x)^4 = 1 + 4*x + 11*x^2 + 27*x^3 + 65*x^4 + O(x^5), giving row 4 as (65, 27, 11, 4, 1). (End)