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
Keywords
Examples
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.
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Haskell
a096494 = (* 2) . a000006 -- Reinhard Zumkeller, Sep 20 2014
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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
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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 *)
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