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%I #16 May 06 2022 13:13:51
%S 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,
%T 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,
%U 6,4,3,2,3,3,4,3,6,3,3,3,3,3,3,3,4,3,6
%N The number of elements in the continued fraction for phi(n)/n, where phi is the Euler totient function (A000010).
%H Amiram Eldar, <a href="/A342866/b342866.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = 2 if and only if n is in A007694.
%F a(p) = 3 for an odd prime p.
%e a(2) = 2 since the continued fraction of phi(2)/2 = 1/2 = 0 + 1/2 has 2 elements: {0, 2}.
%e a(3) = 3 since the continued fraction of phi(3)/3 = 2/3 = 0 + 1/(1 + 1/2) has 3 elements: {0, 1, 2}.
%e 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}.
%t a[n_] := Length @ ContinuedFraction[EulerPhi[n]/n]; Array[a, 100]
%o (PARI) a(n) = #contfrac(eulerphi(n)/n); \\ _Michel Marcus_, Mar 30 2021
%Y Cf. A000010, A007694, A076512, A109395, A342867.
%Y Cf. A071862 (similar, with sigma(n)/n).
%K nonn,easy
%O 1,2
%A _Amiram Eldar_, Mar 27 2021
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