A079613 -id:A079613 - cp's OEIS frontend

A081460 Consider the mapping f(r) = (1/2)*(r + N/r) from rationals to rationals where N = 5. Starting with a = 2 and applying the mapping to each new (reduced) rational number gives 2, 9/4, 161/72, 51841/23184, ..., tending to N^(1/2). Sequence gives values of the denominators.

1, 4, 72, 23184, 2403763488, 25840354427429161536, 2986152136938872067784669198846010266752, 39878504028822311675150039382403961856254569551519724209276629577579916539865344

Offset: 1

Amarnath Murthy, Mar 22 2003

nonn

Related sequence pairs (numerator, denominator) can be obtained by choosing N = 2, 3, 6, etc.

The sequence satisfies the Pell equation A081459(n+1)^2 - 5*a(n+1)^2 = 1. - Vincenzo Librandi, Dec 20 2011

Vincenzo Librandi, Table of n, a(n) for n = 1..11

Cf. A000032, A000045, A000129, A001333, A007283, A079613, A081459.

m:=8; f:=[ n eq 1 select 2 else (Self(n-1)+5/Self(n-1))/2: n in [1..m] ]; [ Denominator(f[n]): n in [1..m] ]; // Bruno Berselli, Dec 20 2011

Table[Fibonacci[2^(n - 1)*3], {n, 1, 8}]/2 (* Amiram Eldar, Apr 07 2023 *)

{r=2; N=5; for(n=1,8,a=numerator(r); b=denominator(r); print1(b,","); r=(1/2)*(r + N/r))}

a(n) = 2*a(n-1)*A081459(n-1). - Mario Catalani (mario.catalani(AT)unito.it), May 21 2003
a(n) = A000045(A007283(n-1))/2. - Ehren Metcalfe, Oct 07 2017
From Amiram Eldar, Apr 07 2023: (Start)
a(n) = A079613(n-1)/2.
a(n) = Product_{k=1..n-1} L(3*2^(n-1-k)), where L(k) is the k-th Lucas number (A000032). (End)
a(n) = A001076(2^(n-1)). - Robert FERREOL, Apr 18 2023

Edited and extended by Klaus Brockhaus and Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Apr 06 2003

a(8) corrected by Vincenzo Librandi, Dec 20 2011

A081976 For n > 2, a(n) is minimal so that the products of two adjacent terms are distinct Fibonacci numbers.

Original entry on oeis.org

1, 2, 4, 36, 1288, 3732552, 13845944773136, 431339599553022278260254864, 184905369551724915055273665254253822188651964997391392

Offset: 1

Amarnath Murthy, Apr 03 2003

nonn

Cf. A081974, A081975, A079613.

a(n-1)*a(n) = A079613(n-2) = F(2^(n-2)*3), where F(k) is the k-th Fibonacci number.

A083523 Smallest Fibonacci number divisible by 2^n.

Original entry on oeis.org

1, 2, 8, 8, 144, 46368, 4807526976, 51680708854858323072, 5972304273877744135569338397692020533504, 79757008057644623350300078764807923712509139103039448418553259155159833079730688

Offset: 0

Amarnath Murthy and Meenakshi Srikanth (menakan_s(AT)yahoo.com), May 05 2003

nonn

The index of the Fibonacci numbers above begin: 1, 3, 6, 6 and then doubles thereafter.

Amiram Eldar, Table of n, a(n) for n = 0..12
Ron Knott, The first 300 Fibonacci numbers, completely factorised.
Tamás Lengyel, The order of the Fibonacci and Lucas numbers, The Fibonacci Quarterly, Vol. 33, No. 3 (1995), pp. 234-239.

Cf. A000045, A001622, A007283, A079613, A337923.

Do[k = 1; While[ !IntegerQ[ Fibonacci[k]/2^n], k++ ]; Print[ Fibonacci[k]], {n, 0, 10}]
With[{fibs=Fibonacci[Range[1000]]},Table[SelectFirst[fibs, Divisible[#,2^n]&],{n,0,10}]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Feb 02 2021 *)
Join[{1, 2, 8}, Table[Fibonacci[3*2^(n - 2)], {n, 3, 9}]] (* Amiram Eldar, Jan 29 2022 *)

From Amiram Eldar, Jan 29 2022: (Start)
a(n) = Fibonacci(3*2^(n-2)) = A000045(A007283(n-2)) = A079613(n-2), for n > 2.
Sum_{n>=0} 1/a(n) = 19/8 - 1/phi, where phi is the golden ratio (A001622). (End)

Edited and extended by Robert G. Wilson v, May 06 2003

cp's OEIS Frontend

A081460 Consider the mapping f(r) = (1/2)*(r + N/r) from rationals to rationals where N = 5. Starting with a = 2 and applying the mapping to each new (reduced) rational number gives 2, 9/4, 161/72, 51841/23184, ..., tending to N^(1/2). Sequence gives values of the denominators.

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A081976 For n > 2, a(n) is minimal so that the products of two adjacent terms are distinct Fibonacci numbers.

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A083523 Smallest Fibonacci number divisible by 2^n.

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