A135318 The Kentucky-2 sequence: a(n) = a(n-2) + 2*a(n-4), with a[0..3] = [1, 1, 1, 2].

1, 1, 1, 2, 3, 4, 5, 8, 11, 16, 21, 32, 43, 64, 85, 128, 171, 256, 341, 512, 683, 1024, 1365, 2048, 2731, 4096, 5461, 8192, 10923, 16384, 21845, 32768, 43691, 65536, 87381, 131072, 174763, 262144, 349525, 524288, 699051, 1048576, 1398101, 2097152, 2796203

Offset

0,4

Comments

Shifted Jacobsthal recurrence.

From L. Edson Jeffery, Apr 21 2011: (Start)

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

U=U_(6,2)=

(0 0 1)

(0 2 0)

(2 0 1),

let i in {0,1}, m>=0 an integer and n=2*m+i. Then a(n)=a(2*m+i)=Sum_{j=0..2} (U^m)_(i,j). (End)

a(n) is also the pebbling number of the cycle graph C_{n+1} for n > 1. - Eric W. Weisstein, Jan 07 2021

From Greg Dresden and Ziyi Xie, Aug 25 2023: (Start)

a(n) is the number of ways to tile a zig-zag strip of n cells using squares (of 1 cell) and triangles (of 3 cells). Here is the zig-zag strip corresponding to n=11, with 11 cells:

___ ___

___| |___| |___

| |___| |___| |___

|___| |___| |___| |

| |___| |___| |___|

|___| |___| |___|,

and here are the two types of triangles (where one is just a reflection of the other):

___ ___

| |___ ___| |

| | | |

| ___| and |___ |

|___| |___|.

As an example, here is one of the a(11) = 32 ways to tile the zig-zag strip of 11 cells:

___ ___

___| |___| |___

| |___| | |___

| |___ | |

| ___| |___| ___|

|___| |___| |___|. (End)

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..5000

Minerva Catral et al., Generalizing Zeckendorf's Theorem: The Kentucky Sequence, arXiv:1409.0488 [math.NT], 2014. See 1.3 p. 2, same sequence without the first 2 terms.

L. E. Jeffery, Unit-primitive matrices.

Eric Weisstein's World of Mathematics, Cycle Graph

Eric Weisstein's World of Mathematics, Pebbling Number

Index entries for linear recurrences with constant coefficients, signature (0,1,0,2).

Formula

From R. J. Mathar, Feb 19 2008: (Start)

O.g.f.: [1/(1+x^2)+(-2-3*x)/(2*x^2-1)]/3.

a(2n) = A001045(n+1).

a(2n+1) = A000079(n). (End)

From L. Edson Jeffery, Apr 21 2011: (Start)

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

a(n) = (((-i)^(n+1)-i^(n+1))*2*i*sqrt(2)+3*(1+(-1)^(n+1))*2^((n+2)/2)+(1-(-1)^(n+1))*2^((n+5)/2))/(12*sqrt(2)), where i=sqrt(-1). (End)

a(n) = (2^floor(n/2)*(5-(-1)^n)+(-1)^floor(n/2)*(1+(-1)^n))/6 = (A016116(n)*A010711(n)+2*A056594(n))/6. - Bruno Berselli, Apr 21 2011

a(2n) = 2*a(2n-1) - a(2n-2); a(2n+1) = a(2n) + a(2n-2). - Richard R. Forberg, Aug 19 2013

a(n) = A112387(n + (-1)^n). - Alois P. Heinz, Sep 28 2023

Example

Let i=0 and m=3. Then U^3 = (2,0,3;0,8,0;6,0,5), and the first-row sum (corresponding to i=0) is 2 + 0 + 3 = 5. Hence a(n) = a(2*m+i) = a(2*3+0) = a(6) = 2 + 3 = 5.

Maple

a:= n-> (<<0|1>, <2|1>>^(iquo(n, 2, 'm')). <<1, 1+m>>)[1, 1]:
seq(a(n), n=0..50);  # Alois P. Heinz, May 30 2022

Mathematica

LinearRecurrence[{0, 1, 0, 2}, {1, 1, 1, 2}, 40] (* Harvey P. Dale, Oct 14 2015 *)

Prog

(Magma) [(2^Floor(n/2)*(5-(-1)^n)+(-1)^Floor(n/2)*(1+(-1)^n))/6: n in [0..50]]; // Vincenzo Librandi, Aug 10 2011

Crossrefs

Cf. A000079, A001045, A112387.

Sequence in context: A302592 A078762 A103262 * A210671 A189761 A329576

Adjacent sequences: A135315 A135316 A135317 * A135319 A135320 A135321

Keyword

nonn,easy

Author

Paul Curtz, Feb 16 2008

Extensions

More terms from R. J. Mathar, Feb 19 2008

Status

approved