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a(n) = Fib(3*n)^2 - 2*Fib(3*n) + 4*Fib(3*n+1) + 5, where Fib = A000045.
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%I #13 Aug 10 2024 22:33:16

%S 9,17,105,1313,21385,374833,6688617,119863873,2150198793,38580909137,

%T 692294282601,12422665840481,222915477614473,4000055027956849,

%U 71778071199324777,1288005210351490177

%N a(n) = Fib(3*n)^2 - 2*Fib(3*n) + 4*Fib(3*n+1) + 5, where Fib = A000045.

%C Continued fraction for sqrt(a(n)) has period n.

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (22, -71, -36, 73, 14, -1).

%F G.f.: (-x^5 - 91*x^4 + 534*x^3 + 370*x^2 - 181*x + 9)/((x^2-18*x+1)*(x^2-1)*(x^2+4*x-1)). - _Michael Somos_

%F a(n) = 22*a(n-1) - 71*a(n-2) - 36*a(n-3) + 73*a(n-4) + 14*a(n-5) - a(n-6).

%t #[[1]]^2-2#[[1]]+4#[[2]]+5&/@Table[{Fibonacci[3n],Fibonacci[3n+1]},{n,0,20}] (* or *) LinearRecurrence[{22,-71,-36,73,14,-1},{9,17,105,1313,21385,374833},20] (* _Harvey P. Dale_, Jul 07 2017 *)

%o (PARI) a(n) = fibonacci(3*n)^2 - 2*fibonacci(3*n) + 4*fibonacci(3*n+1) + 5 \\ _Andrew Howroyd_, Aug 10 2024

%Y Cf. A000045 (Fibonacci numbers), A003285.

%K nonn

%O 0,1

%A _David W. Wilson_