a(n) = smallest prime >a(n-1) such that sum of first n terms is Fib no.

Jack Brennen jack at brennen.net
Sat Dec 21 01:34:17 CET 2002

Neil Fernandez wrote:
> 2,3,29,199,2351,24155233, next term >10^15
> a(n) = smallest prime > a(n-1) such that sum of first n terms is a
> Fibonacci number
> a(7), if it exists, > 10^15

I'll give the Fibonacci indices, because it's a more compact
representation of the sequence...  I'm pretty sure it goes like:

ai(n) = 0,3,5,9,13,18,37,384,569,2760...

(Only "pretty sure" because I haven't rigorously proved the
 numbers primes...)

This sequence is defined:

  ai(0) = 0
  ai(1) = 3
  ai(n+1) = the smallest x such that U(x)-U(ai(n)) is both prime
            and greater than U(ai(n))-U(ai(n-1))

  Where U(x) is the Lucas sequence defined by (P,Q)=(1,-1) --
   the Fibonacci numbers.

The sequence a(n) in the quoted post can be defined for n>=1 as:

  a(n) = U(ai(n))-U(ai(n-1))

a(7), a(8), and a(9) have 80, 119, and 577 digits respectively.


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