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%I #19 Sep 10 2019 03:19:38
%S 1,7,175,3423,65807,1263367,24251423,465522687,8936020191,
%T 171532889927,3292688640591,63205362446303,1213269239181167,
%U 23289515157668039,447057832476812095,8581574336168940799,164729063528963009727
%N Let M be the 3 X 3 matrix [0 1 0 / 0 0 1 / 7 -35 21]. Take M^n * [1 1 1] = [p q r]; then a(n-1), a(n), a(n+1) = -p, -q, -r respectively.
%C a(n)/a(n-1) tends to tan^2(3*Pi/7) = 19.195669358...
%C The matrix M has an eigenvalue of tan^2(3*Pi/7) which is one root of x^3 - 21*x^2 + 35*x - 7 (the two other roots being tan^2(Pi/7) and tan^2(2*Pi/7)).
%D C. V. Durell and A. Robson, "Advanced Trigonometry", Dover 2003, p. 205.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (21,-35,7).
%F G.f.: x*(14*x^3 - 63*x^2 + 14*x - 1)/(7*x^3 - 35*x^2 + 21*x - 1). [_Colin Barker_, Nov 08 2012]
%e a(4), a(5), a(6) = 3423, 65807, 1263367 since M^5 * [1 1 1] = [ -3423 -65807 -1263367].
%t Table[ Abs[ MatrixPower[{{0, 1, 0}, {0, 0, 1}, {7, -35, 21}}, n].{1, 1, 1}][[2]], {n, 17}] (* _Robert G. Wilson v_, Apr 28 2004 *)
%K nonn,easy
%O 1,2
%A _Gary W. Adamson_, Apr 25 2004
%E More terms from _Robert G. Wilson v_, Apr 28 2004
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