A071866 Number of elements in the continued fraction for prime(n+1)/prime(n).

2, 3, 3, 4, 3, 3, 3, 4, 4, 3, 3, 3, 3, 4, 4, 4, 3, 3, 4, 3, 3, 4, 4, 3, 3, 3, 4, 3, 3, 3, 4, 4, 3, 4, 3, 3, 3, 4, 4, 4, 3, 3, 3, 3, 3, 6, 6, 4, 3, 3, 4, 3, 3, 4, 4, 4, 3, 3, 3, 3, 4, 4, 4, 3, 3, 6, 3, 5, 3, 3, 4, 4, 3, 3, 4, 4, 6, 3, 3, 4, 3, 3, 3, 3, 4, 4, 3, 3, 3, 4, 4, 4, 4, 5, 4, 4, 5, 3, 3, 3, 5, 4, 4, 3, 3

Offset: 1

Author

Benoit Cloitre, Jun 09 2002

Keywords

• easy
• nonn

Examples

			prime(5)/prime(4) = 11/7, 11/7 continued fraction is [1, 1, 1, 3] which contains 4 elements, hence a(4)=4.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A110021, A109374, A112323, A112324, A112768.

Programs

Maple

seq(nops(convert(ithprime(n+1)/ithprime(n),confrac)),n=1..200); # Robert Israel, May 29 2018

Mathematica

Table[Length[ContinuedFraction[Prime[n + 1]/Prime[n]]], {n, 105}] (* Ray Chandler, Sep 18 2005 *)

PARI

a(n) = length(contfrac(prime(n+1)/prime(n)));

Extensions

More terms from Hans Havermann, Jul 06 2002