A131402 -id:A131402 - cp's OEIS frontend

A131403 Row sums of triangle A131402.

1, 2, 5, 10, 21, 44, 93, 196, 411, 856, 1771, 3642, 7451, 15178, 30809, 62358, 125921, 253800, 510777, 1026704, 2061751, 4137012, 8295895, 16627190, 33311671, 66716054, 133582133, 267407026, 535206861, 1071049316, 2143127061, 4287918172, 8578528851, 17161414288

Offset: 0

Gary W. Adamson, Jul 07 2007

nonn

			a(4) = 21 = sum of row 4 terms of A131402: (1 + 6 + 7 + 6 + 1).

Andrew Howroyd, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (5, -8, 3, 3, -2)

Cf. A131402.

LinearRecurrence[{5,-8,3,3,-2},{1,2,5,10,21},40] (* Harvey P. Dale, Nov 22 2021 *)

Vec((1 - 3*x + 3*x^2 - 2*x^3 + 2*x^4)/((1 - x)^2*(1 - 2*x)*(1 - x - x^2)) + O(x^40)) \\ Andrew Howroyd, Aug 09 2018

From Andrew Howroyd, Aug 09 2018: (Start)
a(n) = 5*a(n-1) - 8*a(n-2) + 3*a(n-3) + 3*a(n-4) - 2*a(n-5).
G.f.: (1 - 3*x + 3*x^2 - 2*x^3 + 2*x^4)/((1 - x)^2*(1 - 2*x)*(1 - x - x^2)).
(End)

Terms a(10) and beyond from Andrew Howroyd, Aug 09 2018

A131404 a(n) = 2*A131402(n) - 1.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 7, 7, 1, 1, 11, 13, 11, 1, 1, 13, 27, 27, 13, 1, 1, 17, 39, 65, 39, 17, 1, 1, 19, 61, 111, 111, 61, 19, 1, 1, 23, 79, 193, 221, 193, 79, 23, 1, 1, 25, 109, 283, 433, 433, 283, 109, 25, 1, 1, 29, 133, 425, 715, 925, 715, 425, 133, 29, 1

Offset: 0

Gary W. Adamson, Jul 07 2007

Row sums = A131405: (1, 2, 7, 16, 37, 82, 179, ...).

			First few rows of the triangle are:
  1;
  1,  1;
  1,  5,  1;
  1,  7,  7,  1;
  1, 11, 13, 11,  1;
  1, 13, 27, 27, 13,  1;
  1, 17, 39, 65, 39, 17,  1;
  ...

Andrew Howroyd, Table of n, a(n) for n = 0..1274

Row sums are A131405.

Cf. A131402.

/* As triangle */ [[4*Binomial(n, k) + 1 - 2*Binomial(Floor(n + k) div 2, k) - 2*Binomial(n-Floor((k+1)/2), Floor(k/2)): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Aug 10 2018

T(n,k) = if(k <= n, 4*binomial(n,k) + 1 - 2*binomial((n + k)\2, k) - 2*binomial(n-(k+1)\2, k\2), 0) \\ Andrew Howroyd, Aug 09 2018

T(n,k) = 4*binomial(n, k) + 1 - 2*binomial(floor((n + k)/2), k) - 2*binomial(n-floor((k+1)/2), floor(k/2)). - Andrew Howroyd, Aug 09 2018

Terms a(55) and beyond from Andrew Howroyd, Aug 09 2018

cp's OEIS Frontend

A131403 Row sums of triangle A131402.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions