|
|
A145231
|
|
a(n) = Fibonacci(4^n).
|
|
9
|
|
|
|
OFFSET
|
0,2
|
|
COMMENTS
|
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
|
|
LINKS
|
|
|
FORMULA
|
a(n) = (G^(4^n) - (1-G)^(4^n) )/sqrt(5) where G = (1 + sqrt 5)/2 = A001622.
a(n) = round( sqrt(4/5) *cosh( 4^n*arccosh (sqrt(5/4)) )).
a(n+1) = a(n)*(5*a(n)^2 + 2)*sqrt(5*a(n)^2 + 4) for n >= 1.
a(n) == 3 (mod 4) for n >= 1.
a(n+1) == a(n) (mod 2^(2*n+1)).
A341601(n) == a(n) (mod 2^n) for n >= 2.
In the ring of 2-adic integers, the sequence {Fibonacci(4^n)} converges to the 2-adic integer A341603. (End)
|
|
MAPLE
|
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
|
|
MATHEMATICA
|
G = (1 + Sqrt[5])/2; Table[Expand[(G^(4^n) - (1 - G)^(4^n))/Sqrt[5]], {n, 1, 6}]
Table[Round[(4/5)^(1/2)*Cosh[4^n*ArcCosh[((5/4)^(1/2))]]], {n, 1, 7}]
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|