A084130 a(n) = 8*a(n-1) - 8*a(n-2), a(0)=1, a(1)=4.

1, 4, 24, 160, 1088, 7424, 50688, 346112, 2363392, 16138240, 110198784, 752484352, 5138284544, 35086401536, 239584935936, 1635988275200, 11171226714112, 76281907511296, 520885446377472, 3556828310929408, 24287542916415488

Offset

0,2

Comments

Binomial transform of A001541.

Let A be the unit-primitive matrix (see [Jeffery]) A = A_(8,3) = [0,0,0,1; 0,0,2,0; 0,2,0,1; 2,0,2,0]. Then A084130(n) = (1/4)*Trace(A^(2*n)). (Cf. A006012, A001333.) - L. Edson Jeffery, Apr 04 2011

a(n) is also the rational part of the Q(sqrt*(2)) integer giving the length L(n) of a variant of the Lévy C-curve, given by _Kival Ngaokrajan_, at iteration step n. See A057084. - Wolfdieter Lang, Dec 18 2014

Links

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

L. E. Jeffery, Unit-primitive matrices

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

Formula

a(n) = (4+sqrt(8))^n/2 + (4-sqrt(8))^n/2.

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

E.g.f.: exp(4*x)*cosh(sqrt(8)*x).

a(n) = A057084(n) - 4*A057084(n-1). - R. J. Mathar, Nov 10 2013

From G. C. Greubel, Oct 13 2022: (Start)

a(2*n) = 2^(3*n-1)*A002203(2*n).

a(2*n+1) = 2^(3*n+2)*A000129(2*n+1). (End)

Mathematica

LinearRecurrence[{8, -8}, {1, 4}, 30] (* Harvey P. Dale, Sep 25 2014 *)

Prog

(PARI) {a(n)= if(n<0, 0, real((4+ 2*quadgen(8))^n))}
(Magma) [n le 2 select 4^(n-1) else 8*(Self(n-1) -Self(n-2)): n in [1..41]]; // G. C. Greubel, Oct 13 2022
(SageMath)
A084130=BinaryRecurrenceSequence(8, -8, 1, 4)
[A084130(n) for n in range(41)] # G. C. Greubel, Oct 13 2022

Crossrefs

Cf. A000129, A001333, A001541, A002203, A006012, A057084, A084130, A084131.

Sequence in context: A344735 A117337 A272865 * A059304 A069722 A343842

Adjacent sequences: A084127 A084128 A084129 * A084131 A084132 A084133

Keyword

easy,nonn

Author

Paul Barry, May 16 2003

Status

approved