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%I #34 Jul 12 2021 03:37:04
%S 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,
%T 20,20,20,20,22,22,22,22,24,24,24,24,24,26,26,26,26,26,28,28,28,28,30,
%U 30,30,30,30,30,32,32,32,32,32,32,32,34,34,34,34,34,36,36,36,36,36,36
%N Largest value in the periodic part of the continued fraction of sqrt(prime(n)).
%H Reinhard Zumkeller, <a href="/A096494/b096494.txt">Table of n, a(n) for n = 1..10000</a>
%F It seems that lim_{n->infinity} a(n)/n = 0. - _Benoit Cloitre_, Apr 19 2003
%F a(n) = 2*A000006(n). - _Benoit Cloitre_, Apr 19 2003
%e 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.
%p A096491 := proc(n)
%p if issqr(n) then
%p sqrt(n) ;
%p else
%p numtheory[cfrac](sqrt(n),'periodic','quotients') ;
%p %[2] ;
%p max(op(%)) ;
%p end if;
%p end proc:
%p A096494 := proc(n)
%p option remember ;
%p A096491(ithprime(n)) ;
%p end proc: # _R. J. Mathar_, Mar 18 2010
%t {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
%t a[n_]:=IntegerPart[Sqrt[Prime[n]]] 2 IntegerPart[Sqrt[#]]&/@Prime[Range[90]] (* _Vincenzo Librandi_, Aug 09 2015 *)
%o (Haskell)
%o a096494 = (* 2) . a000006 -- _Reinhard Zumkeller_, Sep 20 2014
%Y Cf. A000006, A003285, A005980, A054269.
%Y Cf. A096491, A096492, A096493, A096495, A096496.
%Y Cf. A117767.
%K nonn
%O 1,1
%A _Labos Elemer_, Jun 29 2004
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