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%I #38 Feb 29 2024 01:51:24
%S 1,5,75025,59425114757512643212875125,
%T 18526362353047317310282957646406309593963452838196423660508102562977229905562196608078556292556795045922591488273554788881298750625
%N a(n) = Fibonacci(5^n).
%H Seiichi Manyama, <a href="/A145232/b145232.txt">Table of n, a(n) for n = 0..5</a>
%H Robert Frontczak, <a href="https://www.fq.math.ca/Problems/FQElemProbFeb2024.pdf">Problem B-1341</a>, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 62, No. 1 (2024), p. 84.
%H Thomas Koshy and Zhenguang Gao, <a href="https://www.fq.math.ca/Papers1/55-1/KoshyGao10272016.pdf">Polynomial Extensions of a Diminnie Delight</a>, Fibonacci Quart. 55 (2017), no. 1, 13-20.
%H Achilleas Sinefakopoulos, <a href="https://smc.math.ca/wp-content/uploads/crux-pdfs/Crux_v20n10_Dec.pdf">Solution to Problem 1909</a>, Crux Mathematicorum, 20 (1994), 295-296.
%F a(n) = (G^(5^n) - (1 - G)^(5^n))/sqrt(5) where G = (1 + sqrt(5))/2.
%F a(n) = (2/sqrt(5))*cosh((2*k+1)^n*arccosh(sqrt(5)/2)).
%F a(n) = (2/sqrt(5))*cosh(5^n*arccosh(sqrt(5)/2)).
%F a(n) = (5^n)*A128935(n). - _R. J. Mathar_, Nov 04 2010
%F a(n) = A000045(A000351(n)). - _Michel Marcus_, Nov 07 2013
%F a(n+1) = 25*a(n)^5 - 25*a(n)^3 + 5*a(n) with a(0) = 1. - _Peter Bala_, Nov 24 2022
%F a(n) = 5^n * Product_{k=0..n-1} (5*a(k)^4 - 5*a(k)^2 + 1) (Frontczak, 2024). - _Amiram Eldar_, Feb 29 2024
%p a := proc(n) option remember; if n = 0 then 1 else 25*a(n-1)^5 - 25*a(n-1)^3 + 5*a(n-1) end if; end:
%p seq(a(n), n = 0..5); # _Peter Bala_, Nov 24 2022
%t G = (1 + Sqrt[5])/2; Table[Expand[(G^(5^n) - (1 - G)^(5^n))/Sqrt[5]], {n, 1, 6}]
%t Table[Round[N[(2/Sqrt[5])*Cosh[5^n*ArcCosh[Sqrt[5]/2]], 1000]], {n, 1, 4}]
%t Fibonacci[5^Range[0,4]] (* _Harvey P. Dale_, Nov 29 2018 *)
%Y Cf. A000045.
%Y Cf. (k^n)-th Fibonacci number: A058635 (k=2), A045529 (k=3), A145231 (k=4), this sequence (k=5), A145233 (k=6), A145234 (k=7), A250487 (k=8), A250488 (k=9), A250489 (k=10).
%K nonn,easy
%O 0,2
%A _Artur Jasinski_, Oct 05 2008
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