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%I #40 Dec 04 2022 04:56:25
%S 1,3,987,10610209857723,
%T 141693817714056513234709965875411919657707794958199867
%N a(n) = Fibonacci(4^n).
%C This sequence has the property that a(n+1) is divisible by a(n). Conjecture: each prime divisor can occur only once (i.e., all terms are squarefree). - _Artur Jasinski_, Oct 05 2008
%H Seiichi Manyama, <a href="/A145231/b145231.txt">Table of n, a(n) for n = 0..6</a>
%F a(n) = (G^(4^n) - (1-G)^(4^n) )/sqrt(5) where G = (1 + sqrt 5)/2 = A001622.
%F a(n) = round( sqrt(4/5) *cosh( 4^n*arccosh (sqrt(5/4)) )).
%F a(n)= A000045(A000302(n)). - _Michel Marcus_, Nov 07 2013
%F From _Peter Bala_, Nov 11 2022: (Start)
%F a(n+1) = a(n)*(5*a(n)^2 + 2)*sqrt(5*a(n)^2 + 4) for n >= 1.
%F a(n) == 3 (mod 4) for n >= 1.
%F a(n+1) == a(n) (mod 2^(2*n+1)).
%F A341601(n) == a(n) (mod 2^n) for n >= 2.
%F In the ring of 2-adic integers, the sequence {Fibonacci(4^n)} converges to the 2-adic integer A341603. (End)
%p a := proc(n) option remember; if n = 1 then 3 else a(n-1)*(5*a(n-1)^2 + 2)*sqrt(5*a(n-1)^2 + 4) end if; end: seq(a(n), n = 1..5); # _Peter Bala_, Nov 14 2022
%t G = (1 + Sqrt[5])/2; Table[Expand[(G^(4^n) - (1 - G)^(4^n))/Sqrt[5]], {n, 1, 6}]
%t Table[Round[(4/5)^(1/2)*Cosh[4^n*ArcCosh[((5/4)^(1/2))]]], {n, 1, 7}]
%t Fibonacci[4^Range[5]] (* _Harvey P. Dale_, Mar 28 2012 *)
%Y Cf. A000045, A341601, A341603.
%Y Cf. (k^n)-th Fibonacci number: A058635 (k=2), A045529 (k=3), this sequence (k=4), A145232 (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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