|
%I #14 May 29 2018 21:40:16
%S 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,
%T 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,
%U 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
%N Number of elements in the continued fraction for prime(n+1)/prime(n).
%H Robert Israel, <a href="/A071866/b071866.txt">Table of n, a(n) for n = 1..10000</a>
%e prime(5)/prime(4) = 11/7, 11/7 continued fraction is [1, 1, 1, 3] which contains 4 elements, hence a(4)=4.
%p seq(nops(convert(ithprime(n+1)/ithprime(n),confrac)),n=1..200); # _Robert Israel_, May 29 2018
%t Table[Length[ContinuedFraction[Prime[n + 1]/Prime[n]]], {n, 105}] - _Ray Chandler_, Sep 18 2005
%o (PARI) for(n=1,200,print1(length(contfrac(prime(n+1)/prime(n))),","))
%Y Cf. A110021, A109374, A112323, A112324, A112768.
%K easy,nonn
%O 1,1
%A _Benoit Cloitre_, Jun 09 2002
%E More terms from _Hans Havermann_, Jul 06 2002
|
