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A342866
The number of elements in the continued fraction for phi(n)/n, where phi is the Euler totient function (A000010).
3
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
1,2
LINKS
FORMULA
a(n) = 2 if and only if n is in A007694.
a(p) = 3 for an odd prime p.
EXAMPLE
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}.
MATHEMATICA
a[n_] := Length @ ContinuedFraction[EulerPhi[n]/n]; Array[a, 100]
PROG
(PARI) a(n) = #contfrac(eulerphi(n)/n); \\ Michel Marcus, Mar 30 2021
CROSSREFS
Cf. A071862 (similar, with sigma(n)/n).
Sequence in context: A139713 A171465 A178620 * A023524 A068639 A074070
KEYWORD
nonn,easy
AUTHOR
Amiram Eldar, Mar 27 2021
STATUS
approved