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A059854 Period of continued fraction for sqrt(n^2+5), n >= 3.

%I A059854 #19 Jul 10 2024 02:57:19
%S A059854 4,6,2,3,6,8,10,2,4,9,4,14,2,16,6,12,12,2,16,22,10,24,2,24,12,24,16,2,
%T A059854 6,26,30,26,2,7,20,12,18,2,18,11,20,64,2,20,30,19,22,2,40,20,10,50,2,
%U A059854 10,38,74,14,2,22,64,50,72,2,48,10,30,48,2,22,51,10,36,2,34,12,47,46,2
%N A059854 Period of continued fraction for sqrt(n^2+5), n >= 3.
%C A059854 The old name was "Quotient cycle length of sqrt(n^2+5) for n>=3." - _Jianing Song_, May 01 2021
%H A059854 Amiram Eldar, <a href="/A059854/b059854.txt">Table of n, a(n) for n = 3..10000</a>
%F A059854 If n is a multiple of 5 then a(n) = 2.
%F A059854 a(n) = A003285(n^2+5). - _Jianing Song_, May 01 2021
%e A059854 sqrt(13^2+5) = [13; 5, 4, 5, 26], so a(13) = 4.
%e A059854 sqrt(14^2+5) = [14; 5, 1, 1, 1, 2, 1, 8, 1, 2, 1, 1, 1, 5, 28], so a(14) = 14.
%e A059854 sqrt(15^2+5) = [15; 6, 30], so a(15) = 2.
%e A059854 sqrt(16^2+5) = [16; 6, 2, 3, 7, 1, 3, 1, 2, 1, 3, 1, 7, 3, 2, 6, 32], so a(16) = 16.
%p A059854 with(numtheory): [seq(nops(cfrac(sqrt(k^2+5), 'periodic', 'quotients')[2]), k=3..256)];
%t A059854 a[n_] := Length[ContinuedFraction[Sqrt[n^2 + 5]][[2]]]; Array[a, 100, 3] (* _Amiram Eldar_, Jul 10 2024 *)
%Y A059854 Cf. A003285.
%Y A059854 Period of continued fraction for sqrt(n^2+k): A059853 (k=3), A059855 (k=4), this sequence (k=5).
%K A059854 nonn
%O A059854 3,1
%A A059854 _Labos Elemer_, Feb 27 2001
%E A059854 New name by _Jianing Song_, May 01 2021