A230900 - OEIS

%I #22 Sep 08 2022 08:46:06

%S 4,2,8,16,128,2048,262144,536870912,140737488355328,

%T 75557863725914323419136,10633823966279326983230456482242756608,

%U 803469022129495137770981046170581301261101496891396417650688

%N a(n) = 2^Lucas(n).

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

%C 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.

%C The constant c has various series representations including

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

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

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

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

%H William W. Adams and J. L. Davison, <a href="http://www.jstor.org/stable/2041889">A remarkable class of continued fractions</a>, Proc. Amer. Math. Soc. 65 (1977), 194-198.

%H P. G. Anderson, T. C. Brown, P. J.-S. Shiue, <a href="http://people.math.sfu.ca/~vjungic/tbrown/tom-28.pdf">A simple proof of a remarkable continued fraction identity</a>, Proc. Amer. Math. Soc. 123 (1995), 2005-2009.

%H J. L. Davison, <a href="http://www.jstor.org/stable/2041058">A series and its associated continued fraction</a>, Proc. Amer. Math. Soc. 63 (1977), pp. 29-32.

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

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

%F Sum_{n>=1} 1/a(n) = A121821. - _Amiram Eldar_, Oct 27 2020

%p 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);

%t 2^LucasL[Range[0,15]] (* _Harvey P. Dale_, Jul 21 2015 *)

%o (PARI) for(n=0,10, print1(2^(fibonacci(n+1) + fibonacci(n-1)), ", ")) \\ _G. C. Greubel_, Dec 22 2017

%o (Magma) [2^(Lucas(n)): n in [0..10]]; // _G. C. Greubel_, Dec 22 2017

%Y Cf. A000032, A000301, A014565, A121821.

%K nonn,easy

%O 0,1

%A _Peter Bala_, Oct 31 2013