A083879 a(0)=1, a(1)=4, a(n) = 8*a(n-1) - 14*a(n-2), n >= 2.

1, 4, 18, 88, 452, 2384, 12744, 68576, 370192, 2001472, 10829088, 58612096, 317289536, 1717746944, 9299922048, 50350919168, 272608444672, 1475954689024, 7991119286784, 43265588647936, 234249039168512, 1268274072276992

Offset

0,2

Comments

Binomial transform of A083878.

4th binomial transform of A077957. Inverse binomial transform of A083880. - Philippe Deléham, Nov 30 2008

From L. Edson Jeffery, Apr 26 2011: (Start)

Let G be the Gram matrix

G =

(4 1 0 1)

(1 4 1 0)

(0 1 4 -1)

(1 0 -1 4)

of A028997. Then a(n) = (1/4)*Trace(G^n). (End)

Links

Table of n, a(n) for n=0..21.

Index entries for linear recurrences with constant coefficients, signature (8,-14).

Formula

a(n) = 2^((n-2)/2)*(2*sqrt(2)-1)^n + 2^((n-2)/2)*(2*sqrt(2)+1)^n;

a(n) = Sum_{k=0..n} C(n, 2k)*5^(n-2k)2^k.

G.f.: (1-4x)/(1-8x+14x^2).

E.g.f.: exp(4x)cosh(x*sqrt(2)).

((4+sqrt(2))^n + (4-sqrt(2))^n)/2. Offset=0. a(3)=88. - Al Hakanson (hawkuu(AT)gmail.com), Oct 15 2008

a(n) = Sum_{k=0..n} A098158(n,k)*2^(3*k-n). - Philippe Deléham, Nov 30 2008

Mathematica

LinearRecurrence[{8, -14}, {1, 4}, 30] (* Harvey P. Dale, May 08 2013 *)

Crossrefs

Cf. A028997, A083880.

Sequence in context: A199309 A083325 A050146 * A363184 A081671 A244785

Adjacent sequences: A083876 A083877 A083878 * A083880 A083881 A083882

Keyword

easy,nonn

Author

Paul Barry, May 08 2003

Status

approved