A087131 a(n) = 2^n*Lucas(n), where Lucas = A000032.

2, 2, 12, 32, 112, 352, 1152, 3712, 12032, 38912, 125952, 407552, 1318912, 4268032, 13811712, 44695552, 144637952, 468058112, 1514668032, 4901568512, 15861809152, 51329892352, 166107021312, 537533612032, 1739495309312

Offset: 0

Author

Paul Barry, Aug 16 2003

Keywords

• easy
• nonn

Comments

Number of ways to tile an n-bracelet with two types of colored squares and four types of colored dominoes.

Inverse binomial transform of even Lucas numbers (A014448).

From L. Edson Jeffery, Apr 25 2011: (Start)

Let A be the unit-primitive matrix (see [Jeffery])

A=A_(10,4)=

(0 0 0 0 1)

(0 0 0 2 0)

(0 0 2 0 1)

(0 2 0 2 0)

(2 0 2 0 1).

Then a(n)=(Trace(A^n)-1)/2. Also a(n)=Trace((2*A_(5,1))^n), where A_(5,1)=[(0,1); (1,1)] is also a unit-primitive matrix. (End)

Also the number of connected dominating sets in the n-sun graph for n >= 3. - Eric W. Weisstein, May 02 2017

Also the number of total dominating sets in the n-sun graph for n >= 3. - Eric W. Weisstein, Apr 27 2018

References

• Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, identity 237, p. 132.

Links

• L. Edson Jeffery, Unit-primitive matrices
• Eric Weisstein's World of Mathematics, Connected Dominating Set.
• Eric Weisstein's World of Mathematics, Sun Graph.
• Eric Weisstein's World of Mathematics, Total Dominating Set.
• Index entries for linear recurrences with constant coefficients, signature (2,4).

Crossrefs

First differences of A006483 and A103435.

Cf. A000032, A014334, A014335, A084057, A269992.

Programs

Magma

[2] cat [2^n*Lucas(n): n in [1..30]]; // G. C. Greubel, Dec 18 2017

Mathematica

Table[Tr[MatrixPower[{{2, 2}, {2, 0}}, x]], {x, 1, 20}] (* Artur Jasinski, Jan 09 2007 *)
Join[{2}, Table[2^n LucasL[n], {n, 20}]] (* Eric W. Weisstein, May 02 2017 *)
Join[{2}, 2^# LucasL[#] & [Range[20]]] (* Eric W. Weisstein, May 02 2017 *)
LinearRecurrence[{2, 4}, {2, 12}, {0, 20}] (* Eric W. Weisstein, Apr 27 2018 *)
CoefficientList[Series[(2 (-1 + x))/(-1 + 2 x + 4 x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Apr 27 2018 *)

PARI

for(n=0,30, print1(if(n==0, 2, 2^n*(fibonacci(n+1) + fibonacci(n-1))), ", ")) \\ G. C. Greubel, Dec 18 2017

PARI

first(n) = Vec(2*(1-x)/(1-2*x-4*x^2) + O(x^n)) \\ Iain Fox, Dec 19 2017

Sage

[lucas_number2(n,2,-4) for n in range(0, 25)] # Zerinvary Lajos, Apr 30 2009

Formula

a(n) = 2*A084057(n).

Recurrence: a(n) = 2a(n-1) + 4a(n-2), a(0)=2, a(1)=2.

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

a(n) = (1+sqrt(5))^n + (1-sqrt(5))^n.

For n>=2, a(n) = Trace of matrix [({2,2},{2,0})^n]. - Artur Jasinski, Jan 09 2007

a(n) = 2*[A063727(n)-A063727(n-1)]. - R. J. Mathar, Nov 16 2007

a(n) = (5*A052899(n)-1)/2. - L. Edson Jeffery, Apr 25 2011

a(n) = [x^n] ( 1 + x + sqrt(1 + 2*x + 5*x^2) )^n for n >= 1. - Peter Bala, Jun 23 2015

Sum_{n>=1} 1/a(n) = (1/2) * A269992. - Amiram Eldar, Nov 17 2020

From Amiram Eldar, Jan 15 2022: (Start)

a(n) == 2 (mod 10).

a(n) = 5 * A014334(n) + 2.

a(n) = 10 * A014335(n) + 2. (End)

Extensions

Edited by Ralf Stephan, Feb 08 2005