A096494 Largest value in the periodic part of the continued fraction of sqrt(prime(n)).

2, 2, 4, 4, 6, 6, 8, 8, 8, 10, 10, 12, 12, 12, 12, 14, 14, 14, 16, 16, 16, 16, 18, 18, 18, 20, 20, 20, 20, 20, 22, 22, 22, 22, 24, 24, 24, 24, 24, 26, 26, 26, 26, 26, 28, 28, 28, 28, 30, 30, 30, 30, 30, 30, 32, 32, 32, 32, 32, 32, 32, 34, 34, 34, 34, 34, 36, 36, 36, 36, 36, 36

Offset

1,1

Links

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Formula

It seems that lim_{n->infinity} a(n)/n = 0. - Benoit Cloitre, Apr 19 2003

a(n) = 2*A000006(n). - Benoit Cloitre, Apr 19 2003

Example

n=31: prime(31) = 127, and the periodic part is {3,1,2,2,7,11,7,2,2,1,3,22}, so a(31)=22.

Maple

A096491 := proc(n)
if issqr(n) then
sqrt(n) ;
else
numtheory[cfrac](sqrt(n), 'periodic', 'quotients') ;
%[2] ;
max(op(%)) ;
end if;
end proc:
A096494 := proc(n)
option remember ;
A096491(ithprime(n)) ;
end proc: # R. J. Mathar, Mar 18 2010

Mathematica

{te=Table[0, {m}], u=1}; Do[s=Max[Last[ContinuedFraction[Prime[n]^(1/2)]]]; te[[u]]=s; u=u+1, {n, 1, m}]; te
a[n_]:=IntegerPart[Sqrt[Prime[n]]] 2 IntegerPart[Sqrt[#]]&/@Prime[Range[90]] (* Vincenzo Librandi, Aug 09 2015 *)

Prog

(Haskell)
a096494 = (* 2) . a000006  -- Reinhard Zumkeller, Sep 20 2014

Crossrefs

Cf. A000006, A003285, A005980, A054269.

Cf. A096491, A096492, A096493, A096495, A096496.

Cf. A117767.

Sequence in context: A079584 A179291 A004079 * A116568 A239933 A061106

Adjacent sequences: A096491 A096492 A096493 * A096495 A096496 A096497

Keyword

nonn

Author

Labos Elemer, Jun 29 2004

Status

approved