A160683 -id:A160683 - cp's OEIS frontend

A160686 Numbers n such that n/A000005(A000045(n)) is an integer.

1, 2, 4, 8, 16, 32, 64

Offset: 1

Ctibor O. Zizka, May 23 2009

The first power of 2 not in this sequence is 2^7 = 128, because 128/A000005(A000045(128)) = 1/2, which is not an integer. - Nathaniel Johnston, May 08 2011

Next term, if it exists, is greater than 3000. Conjecture: the sequence is finite and complete. - Max Alekseyev, May 21 2011

Cf. A000005, A000045, A000079, A160683.

with(combinat):with(numtheory): A160686 := proc(n) option remember: local k: if(n=1)then return 1:fi: for k from procname(n-1)+1 do if(k mod tau(fibonacci(k))=0)then return k:fi: od: end: seq(A160686(n),n=1..7); # Nathaniel Johnston, May 08 2011

{n: A063375(n) | n}. - R. J. Mathar, May 25 2009

Inverted division in the definition - R. J. Mathar, May 25 2009

Erroneous term a(5) = 12 removed by Nathaniel Johnston, May 08 2011

A272331 Refactorable Fibonacci numbers.

Original entry on oeis.org

1, 2, 8, 46368, 4807526976

Offset: 1

Waldemar Puszkarz, Apr 25 2016

Luca & Young prove that there are no more terms in this sequence. - Charles R Greathouse IV, Apr 27 2016

			8 is a term as a Fibonacci number that is divisible by the number of its divisors, (1,2,4,8), which is 4.

Florian Luca and Paul Thomas Young, On the number of divisors of n! and of the Fibonacci numbers

Intersection of A000045 (Fibonacci numbers) and A033950 (refactorable numbers).

Cf. A000005 (number of divisors), A160683.

select(t -> t mod numtheory:-tau(t) = 0, map(combinat:-fibonacci, [$2..200])); # Robert Israel, Apr 27 2016

DeleteDuplicates@Select[Fibonacci@Range@200, Divisible[#, IntegerLength@#]&]

for(n=2, 200, fn=fibonacci(n); fn%numdiv(fn)==0&&print1(fn ", "))

a(n) = A000045(A160683(n+1)). - Michel Marcus, Apr 25 2016

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A160686 Numbers n such that n/A000005(A000045(n)) is an integer.

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A272331 Refactorable Fibonacci numbers.

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