cp's OEIS Frontend

A079613 a(n) = F(3*2^n) where F(k) denotes the k-th Fibonacci number.

%I A079613 #24 Jan 05 2025 19:51:37
%S A079613 2,8,144,46368,4807526976,51680708854858323072,
%T A079613 5972304273877744135569338397692020533504,
%U A079613 79757008057644623350300078764807923712509139103039448418553259155159833079730688
%N A079613 a(n) = F(3*2^n) where F(k) denotes the k-th Fibonacci number.
%C A079613 Let b = sqrt(5)/5. We have the alternating series identity (10 - 4*sqrt(5))/5 = b/2 - b^2/(2*8) + b^3/(2*8*144) - b^4/(2*8*144*46368) + ..., so this sequence is a generalized Pierce expansion of (10 - 4*sqrt(5))/5 to the base b as defined in A058635. - _Peter Bala_, Nov 04 2013
%D A079613 Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete mathematics, second edition, Addison Wesley, 1994, p. 557, ex. 6.61.
%H A079613 Amiram Eldar, <a href="/A079613/b079613.txt">Table of n, a(n) for n = 0..10</a>
%H A079613 V. E. Hoggatt, Jr. and Marjorie Bicknell, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/14-5/hoggatt2.pdf">A Reciprocal Series of Fibonacci Numbers with Subscripts 2^n k</a>, The Fibonacci Quarterly, Vol. 14, No. 5 (1976), p. 453-455.
%F A079613 Sum_{n>=0} 1/a(n) = 5/4 - 1/phi = 0.6319660112... since Sum_{k=0..n} 1/a(k) = 5/4 - F(3*2^n-1)/F(3*2^n).
%F A079613 a(n) = A081976(n+1)*A081976(n+2). - _Amarnath Murthy_, Apr 03 2003
%F A079613 a(n) = (1/sqrt(5))*( (2 + sqrt(5))^2^n - 1/(2 + sqrt(5))^2^n ) for n >= 1. - _Peter Bala_, Nov 04 2013
%F A079613 a(n) = A000045(A007283(n)). - _Amiram Eldar_, Jan 29 2022
%t A079613 Table[Fibonacci[3*2^n], {n, 0, 7}] (* _Amiram Eldar_, Jan 29 2022 *)
%o A079613 (Magma) [Fibonacci(3*2^n) : n in [0..7]]; // _Wesley Ivan Hurt_, Apr 05 2023
%Y A079613 Cf. A000045, A001622 (phi), A007283, A058635, A081976, A083523.
%K A079613 nonn
%O A079613 0,1
%A A079613 _Benoit Cloitre_, Jan 29 2003