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

1,2

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).

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

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 *)

Prog

(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 */

Crossrefs

Sequence in context: A129490 A018132 A329758 * A319437 A270659 A159288

Adjacent sequences: A100635 A100636 A100637 * A100639 A100640 A100641

Keyword

easy,nonn

Author

Simone Severini, Dec 04 2004

Extensions

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

Status

approved