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

%I A100638 #21 Mar 03 2024 14:10:31
%S A100638 1,2,3,4,7,10,15,22,37,54,81,118,199,290,435,634,1069,1558,2337,3406,
%T A100638 5743,8370,12555,18298,30853,44966,67449,98302,165751,241570,362355,
%U A100638 528106,890461,1297782,1946673,2837134,4783807,6972050,10458075
%N A100638 Successive powers of the matrix A=[1,2;3,4] written by rows in groups of 4.
%C A100638 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, ...
%H A100638 Alois P. Heinz, <a href="/A100638/b100638.txt">Table of n, a(n) for n = 1..1000</a>
%H A100638 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (0, 0, 0, 5, 0, 0, 0, 2).
%F A100638 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
%F A100638 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
%p A100638 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
%t A100638 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_ *)
%o A100638 (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 */
%K A100638 easy,nonn
%O A100638 1,2
%A A100638 _Simone Severini_, Dec 04 2004
%E A100638 Edited by _Benoit Jubin_, _M. F. Hasler_ and _N. J. A. Sloane_, Dec 01 2008