0,1

Compare with A000301(n) = 2^Fibonacci(n).

The sequence a(n) for n >= 1 gives the sequence of partial quotients (other than the first) in the continued fraction expansion of the transcendental real constant c := sum {n >= 1} 1/2^floor(n*(5 + sqrt(5))/2) = 0.13385 44229 67609 80592 ... = 1/(7 + 1/(2 + 1/(8 + 1/(16 + 1/(128 + 1/(2048 + ...)))))). See Adams Davison 1977. Cf. A014565.

The constant c has various series representations including

c = 1 - sum {n >= 1} 1/2^floor(n*(5 - sqrt(5))/2),

c = sum {n >= 1} floor(n*(5 - sqrt(5))/10)/2^n,

c = 3 - sum {n >= 1} 1/2^floor(n*(15 - sqrt(5))/22) and

c = sum {n >= 1} 1/2^floor(n*(15 + sqrt(5))/22) - 2.

William W. Adams and J. L. Davison, A remarkable class of continued fractions, Proc. Amer. Math. Soc. 65 (1977), 194-198.

P. G. Anderson, T. C. Brown, P. J.-S. Shiue, A simple proof of a remarkable continued fraction identity, Proc. Amer. Math. Soc. 123 (1995), 2005-2009.

J. L. Davison, A series and its associated continued fraction, Proc. Amer. Math. Soc. 63 (1977), pp. 29-32.

a(n) = 2^Lucas(n) = 2^A000032(n).

Recurrence: a(n) = a(n-1)*a(n-2) with a(0) = 4, a(1) = 2.

Sum_{n>=1} 1/a(n) = A121821. - Amiram Eldar, Oct 27 2020

a := proc(n) option remember; if n = 0 then 4 elif n = 1 then 2 else a(n-1)*a(n-2); fi; end; seq(a(n), n = 0..10);

2^LucasL[Range[0, 15]] (* Harvey P. Dale, Jul 21 2015 *)

(PARI) for(n=0, 10, print1(2^(fibonacci(n+1) + fibonacci(n-1)), ", ")) \\ G. C. Greubel, Dec 22 2017

(Magma) [2^(Lucas(n)): n in [0..10]]; // G. C. Greubel, Dec 22 2017

Cf. A000032, A000301, A014565, A121821.

Sequence in context: A386969 A110622 A130078 * A204449 A172393 A245340

Adjacent sequences: A230897 A230898 A230899 * A230901 A230902 A230903

nonn,easy

Peter Bala, Oct 31 2013

approved