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A249578 List of triples (r,s,t): the matrix M = [[4,12,9][2,7,6][1,4,4]] is raised to successive powers, then (r,s,t) are the square roots of M[3,1], M[1,1], M[1,3] respectively.

%I A249578 #16 Sep 08 2022 08:46:10
%S A249578 0,1,0,1,2,3,4,7,12,15,26,45,56,97,168,209,362,627,780,1351,2340,2911,
%T A249578 5042,8733,10864,18817,32592,40545,70226,121635,151316,262087,453948,
%U A249578 564719,978122,1694157,2107560,3650401,6322680
%N A249578 List of triples (r,s,t): the matrix M = [[4,12,9][2,7,6][1,4,4]] 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 A249578 M is the 'Fibonacci matrix' F = [[1,2,1][1,1,0][1,0,0]]  taken to the third power and flipped on a vertical axis.
%C A249578 Sequence identities:
%C A249578 2a(3n-2) + a(3n-1) = a(3n+1)
%C A249578 2a(3n) + a(3n+1) = a(3n+3)
%C A249578 a(3n-2) + a(3n-1) + a(3n+1) = a(3n+2)
%C A249578 a(3n) + a(3n+1) + a(3n-3) = a(3n+2)
%C A249578 a(3n-1) * a(3n) + a(3n+1) * a(3n-2) = a(6n-2).
%H A249578 Colin Barker, <a href="/A249578/b249578.txt">Table of n, a(n) for n = 0..1000</a>
%H A249578 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,4,0,0,-1).
%F A249578 a(n) = 4*a(n-3)-a(n-6).
%F A249578 G.f.: x*(3*x^4-2*x^3+x^2+1) / (x^6-4*x^3+1). - _Colin Barker_, Nov 04 2014
%e A249578 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 A249578 M^1 = [[4,12,9][2,7,6][1,4,4]], so r = a(3) = 1; s = a(4) = 2; t = a(5) = 3.
%t A249578 CoefficientList[Series[x (3 x^4 - 2 x^3 + x^2 + 1) / (x^6 - 4 x^3 + 1), {x, 0, 70}], x] (* _Vincenzo Librandi_, Nov 04 2014 *)
%t A249578 LinearRecurrence[{0,0,4,0,0,-1},{0,1,0,1,2,3},40] (* _Harvey P. Dale_, Jan 17 2017 *)
%o A249578 (PARI) concat(0, Vec(x*(3*x^4-2*x^3+x^2+1)/(x^6-4*x^3+1) + O(x^100))) \\ _Colin Barker_, Nov 04 2014
%o A249578 (Magma) I:=[0,1,0,1,2,3]; [n le 6 select I[n] else 4*Self(n-3)-Self(n-6): n in [1..40]]; // _Vincenzo Librandi_, Nov 04 2014
%Y A249578 a(3n) = the n-th term of A001353.
%Y A249578 a(3n+1) = n-th term of A001075.
%Y A249578 a(3n+2) = n-th term of A005320.
%K A249578 nonn,easy
%O A249578 0,5
%A A249578 _Russell Walsmith_, Nov 03 2014