A342866 The number of elements in the continued fraction for phi(n)/n, where phi is the Euler totient function (A000010).
1, 2, 3, 2, 3, 2, 3, 2, 3, 3, 3, 2, 3, 3, 4, 2, 3, 2, 3, 3, 4, 3, 3, 2, 3, 3, 3, 3, 3, 4, 3, 2, 6, 3, 5, 2, 3, 3, 6, 3, 3, 3, 3, 3, 4, 3, 3, 2, 3, 3, 6, 3, 3, 2, 5, 3, 6, 3, 3, 4, 3, 3, 4, 2, 7, 4, 3, 3, 6, 4, 3, 2, 3, 3, 4, 3, 6, 3, 3, 3, 3, 3, 3, 3, 4, 3, 6
Offset: 1
Examples
a(2) = 2 since the continued fraction of phi(2)/2 = 1/2 = 0 + 1/2 has 2 elements: {0, 2}.
a(3) = 3 since the continued fraction of phi(3)/3 = 2/3 = 0 + 1/(1 + 1/2) has 3 elements: {0, 1, 2}.
a(15) = 4 since the continued fraction of phi(15)/15 = 8/15 = 0 + 1/(1 + 1/(1 + 1/7)) has 4 elements: {0, 1, 1, 7}.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
a[n_] := Length @ ContinuedFraction[EulerPhi[n]/n]; Array[a, 100]
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PARI
a(n) = #contfrac(eulerphi(n)/n); \\ Michel Marcus, Mar 30 2021
Formula
a(n) = 2 if and only if n is in A007694.
a(p) = 3 for an odd prime p.