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List of triples (r,s,t): the matrix M = [[1,4,4][1,3,2][1,2,1]] is raised to successive negative powers, then (r,s,t) are the square roots of M[3,1], M[1,1], M[1,3] respectively.
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%I #21 Aug 02 2024 17:44:19

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

%T -816,577,-408,1970,-1393,985,-4756,3363,-2378,11482,-8119,5741,

%U -27720,19601,-13860,66922,-47321,33461,-161564,114243,-80782,390050,-275807,195025,-941664,665857,-470832

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

%C The sequence comprises, in reverse order, numbers to the right of a(0) in A249576.

%H Colin Barker, <a href="/A249577/b249577.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^4+x^2-x+2) / (x^6-2*x^3-1). - _Colin Barker_, Nov 02 2014

%e M^-1 = [[1,-4,4][-1,3,-2][1,-2,1]]. sqrt(M[1,3]) = 2; M[3,3] = M[1,1] = -1; M[3,1] = 1. Hence a(0) = 2; a(1) = -1; a(2) = 1.

%t LinearRecurrence[{0,0,-2,0,0,1},{2,-1,1,-4,3,-2},50] (* _Harvey P. Dale_, Aug 02 2024 *)

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

%Y Cf. A249576, A000129, A001333, A163271, A077985.

%K sign,tabf,easy

%O 0,1

%A _Russell Walsmith_, Nov 01 2014