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

2, 8, 144, 46368, 4807526976, 51680708854858323072, 5972304273877744135569338397692020533504, 79757008057644623350300078764807923712509139103039448418553259155159833079730688

Offset: 0

Author

Benoit Cloitre, Jan 29 2003

Keywords

• nonn

Comments

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

References

• Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete mathematics, second edition, Addison Wesley, 1994, p. 557, ex. 6.61.

Links

• Amiram Eldar, Table of n, a(n) for n = 0..10
• V. E. Hoggatt, Jr. and Marjorie Bicknell, A Reciprocal Series of Fibonacci Numbers with Subscripts 2^n k, The Fibonacci Quarterly, Vol. 14, No. 5 (1976), p. 453-455.

Crossrefs

Cf. A000045, A001622 (phi), A007283, A058635, A081976, A083523.

Programs

Magma

[Fibonacci(3*2^n) : n in [0..7]]; // Wesley Ivan Hurt, Apr 05 2023

Mathematica

Table[Fibonacci[3*2^n], {n, 0, 7}] (* Amiram Eldar, Jan 29 2022 *)

Formula

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).

a(n) = A081976(n+1)*A081976(n+2). - Amarnath Murthy, Apr 03 2003

a(n) = (1/sqrt(5))*( (2 + sqrt(5))^2^n - 1/(2 + sqrt(5))^2^n ) for n >= 1. - Peter Bala, Nov 04 2013

a(n) = A000045(A007283(n)). - Amiram Eldar, Jan 29 2022