A083881 a(n) = 6*a(n-1) - 6*a(n-2), with a(0)=1, a(1)=3.

1, 3, 12, 54, 252, 1188, 5616, 26568, 125712, 594864, 2814912, 13320288, 63032256, 298271808, 1411437312, 6678993024, 31605334272, 149558047488, 707716279296, 3348949390848, 15847398669312, 74990695670784, 354859782008832

Offset

0,2

Comments

Binomial transform of A001075.

Links

G. C. Greubel, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (6,-6).

Formula

a(n) = ((3-sqrt(3))^n + (3+sqrt(3))^n)/2.

a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k)*3^(n-k).

G.f.: (1-3*x)/(1-6*x+6*x^2).

E.g.f.: exp(3*x) * cosh(x*sqrt(3)).

a(n) = right and left terms in M^n * [1 1 1] where M = the 3X3 matrix [1 1 1 / 1 4 1 / 1 1 1]. M^n * [1 1 1] = [a(n) A030192(n) a(n)]. E.g. a(3) = 54 since M^3 * [1 1 1] = [54 144 54] = [a(3) A030192(3) a(3)]. - Gary W. Adamson, Dec 18 2004

a(n) = Sum_{k, 0<=k<=n}3^k*A098158(n,k). - Philippe Deléham, Dec 04 2006

G.f.: A(x) = G(0) where G(k) = 1 + 3*x/((1-3*x) - x*(1-3*x)/(x + (1-3*x)/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Dec 29 2012.

Mathematica

f[n_]:= Simplify[(3 + Sqrt@3)^n + (3 - Sqrt@3)^n]/2; Array[f, 30, 0] (* Robert G. Wilson v, Oct 31 2010 *)
LinearRecurrence[{6, -6}, {1, 3}, 30] (* G. C. Greubel, Aug 01 2019 *)

Prog

(PARI) my(x='x+O('x^30)); Vec((1-3*x)/(1-6*x+6*x^2)) \\ G. C. Greubel, Aug 01 2019
(Magma) I:=[1, 3]; [n le 2 select I[n] else 6*Self(n-1) -6*Self(n-2): n in [1..30]]; // G. C. Greubel, Aug 01 2019
(Sage) ((1-3*x)/(1-6*x+6*x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Aug 01 2019
(GAP) a:=[1, 3];; for n in [3..30] do a[n]:=6*a[n-1]-6*a[n-2]; od; a; # G. C. Greubel, Aug 01 2019

Crossrefs

Cf. A083882.

Cf. A030192.

Sequence in context: A151205 A151206 A151207 * A151208 A055835 A366118

Adjacent sequences: A083878 A083879 A083880 * A083882 A083883 A083884

Keyword

easy,nonn

Author

Paul Barry, May 08 2003

Status

approved