A059855 Period of continued fraction for sqrt(n^2+4), n >= 1.
1, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2
Offset: 1
Examples
For even n, sqrt(n^2+4) = [n; n/2, 2*n], hence a(n) = 2. For odd n > 1, sqrt(n^2+4) = [n; (n-1)/2, 1, 1, (n-1)/2, 2*n], hence a(n) = 5.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Maple
with(numtheory): [seq(nops(cfrac(sqrt(k^2+4), 'periodic', 'quotients')[2]), k=1..100)];
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Mathematica
a[n_] := Length @ ContinuedFraction[Sqrt[n^2 + 4]][[2]]; Array[a, 100] (* Amiram Eldar, May 13 2020 *)
Formula
a(n) = 2 for even n, a(n) = 5 for odd n > 1.
a(n) = A003285(n^2+4). - Jianing Song, May 01 2021
Comments