%I #22 Dec 29 2021 12:32:01

%S 0,1,0,1,1,2,2,3,4,5,7,10,12,17,24,29,41,58,70,99,140,169,239,338,408,

%T 577,816,985,1393,1970,2378,3363,4756,5741,8119,11482,13860,19601,

%U 27720,33461,47321,66922,80782,114243,161564,195025,275807,390050,470832,665857,941664

%N List of triples (r,s,t): the matrix M = [[1,4,4][1,3,2][1,2,1]] is raised to successive powers, then (r,s,t) are the square roots of M[3,1], M[1,1], M[1,3] respectively.

%C Numbers to the left of a(0) are in A249577.

%C Some identities:

%C a(3n - 2) + a(3n - 1) = a(3n + 1).

%C a(3n) + a(3n + 1) = a(3(n + 1)).

%C a(3n - 2) + a(3n + 1) = a(3n + 2).

%C a(3n) + a(3n - 1) + a(3(n - 2)) = a(3n + 1).

%C a(3n - 1)a(3n) + a(3n + 2)a(3(n + 1)) = a(6n + 2).

%H Colin Barker, <a href="/A249576/b249576.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,2,0,0,1).

%F a(n) = -2*a(n-3)+a(n-6); G.f.: -x*(2*x^4-x^3+x^2+1) / (x^6+2*x^3-1). - _Colin Barker_, Nov 02 2014

%e M^0 = the 3 X 3 identity matrix = [[1,0,0][0,1,0][0,0,1]]. M[3,1] = 0; M[1,1] = 1; M[1,3] = 0. So the first triple is r = a(0) = 0; s = a(1) = 1; t = a(2) = 0.

%e M^1 = [[1,4,4][1,3,2][1,2,1], so r = a(3) = 1; s = a(4) = 1; t = a(5) = 2.

%t LinearRecurrence[{0,0,2,0,0,1},{0,1,0,1,1,2},60] (* _Harvey P. Dale_, Dec 29 2021 *)

%o (PARI) concat(0, Vec(-x*(2*x^4-x^3+x^2+1)/(x^6+2*x^3-1) + O(x^100))) \\ _Colin Barker_, Nov 02 2014

%Y a(3n) = the n-th term of A000129, the Pell numbers.

%Y a(3n+1) = n-th term of A001333.

%Y a(3n+2) = n-th term of A163271.

%K nonn,tabf,easy

%O 0,6

%A _Russell Walsmith_, Nov 01 2014