login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

The number of elements in the continued fraction for phi(n)/n, where phi is the Euler totient function (A000010).
3

%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