A100638 Successive powers of the matrix A=[1,2;3,4] written by rows in groups of 4.

1, 2, 3, 4, 7, 10, 15, 22, 37, 54, 81, 118, 199, 290, 435, 634, 1069, 1558, 2337, 3406, 5743, 8370, 12555, 18298, 30853, 44966, 67449, 98302, 165751, 241570, 362355, 528106, 890461, 1297782, 1946673, 2837134, 4783807, 6972050, 10458075

Offset: 1

Author

Simone Severini, Dec 04 2004

Keywords

• easy
• nonn

Comments

Consider the matrix A = [1, 2; 3, 4]. Then the sequence gives a(1) = A_{1,1} = A_11, a(2) = A_12, a(3) = A_21, a(4) = A_22, a(5)=(A^2)_11, a(6)=(A^2)_12, a(7)=(A^2)_21, a(8)=(A^2)_22, a(9)=(A^3)_11, a(10)=(A^3)_12, ...

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 5, 0, 0, 0, 2).

Programs

Maple

a:= proc(n) local r, m; (<<1|2>, <3|4>>^iquo(n+3, 4, 'r'))[iquo(r+2, 2, 'm'), m+1] end: seq(a(n), n=1..50); # Alois P. Heinz, Dec 01 2008

Mathematica

LinearRecurrence[{0, 0, 0, 5, 0, 0, 0, 2}, {1, 2, 3, 4, 7, 10, 15, 22}, 50] (* Jean-François Alcover, May 18 2018, after R. J. Mathar *)

PARI

A100638(n)=([1,2;3,4]^((n-1)\4+1))[(n-1)%4\2+1,2-n%2] /* M. F. Hasler, Dec 01 2008 */

Formula

a(4n-3) = A124610(n), a(4n-2) = 2 A015535(n), a(4n-1) = 3 A015535(n), a(4n) = a(4n-3) + a(4n-1). - M. F. Hasler, Dec 01 2008

a(n) = 5*a(n-4)+2*a(n-8). a(4n+1)=A124610(n+1), n>=0. G.f.: x*(1+2x+3x^2+4x^3+2x^4+2x^7) / (1-5x^4-2x^8). - R. J. Mathar, Dec 04 2008

Extensions

Edited by Benoit Jubin, M. F. Hasler and N. J. A. Sloane, Dec 01 2008