A049559 a(n) = gcd(n - 1, phi(n)).
1, 1, 2, 1, 4, 1, 6, 1, 2, 1, 10, 1, 12, 1, 2, 1, 16, 1, 18, 1, 4, 1, 22, 1, 4, 1, 2, 3, 28, 1, 30, 1, 4, 1, 2, 1, 36, 1, 2, 1, 40, 1, 42, 1, 4, 1, 46, 1, 6, 1, 2, 3, 52, 1, 2, 1, 4, 1, 58, 1, 60, 1, 2, 1, 16, 5, 66, 1, 4, 3, 70, 1, 72, 1, 2, 3, 4, 1, 78, 1, 2, 1, 82, 1, 4, 1, 2, 1, 88, 1, 18, 1, 4
Offset: 1
Keywords
Examples
a(9) = 2 because phi(9) = 6 and gcd(8, 6) = 2. a(10) = 1 because phi(10) = 4 and gcd(9, 4) = 1.
References
- Richard K. Guy, Unsolved Problems in Number Theory, B37.
Links
- Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
- Eric Weisstein's World of Mathematics, Lehmer's Totient Problem
Crossrefs
Programs
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Magma
[Gcd(n-1, EulerPhi(n)): n in [1..80]]; // Vincenzo Librandi, Oct 13 2018
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Maple
seq(igcd(n-1, numtheory:-phi(n)), n=1..100); # Robert Israel, Nov 09 2015
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Mathematica
Table[GCD[n - 1, EulerPhi[n]], {n, 93}] (* Michael De Vlieger, Nov 09 2015 *)
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PARI
a(n)=gcd(eulerphi(n),n-1) \\ Charles R Greathouse IV, Dec 09 2013
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Python
from sympy import totient, gcd print([gcd(totient(n), n - 1) for n in range(1, 101)]) # Indranil Ghosh, Mar 27 2017
Formula
a(p^m) = a(p) = p - 1 for prime p and m > 0. - Thomas Ordowski, Dec 10 2013
From Antti Karttunen, Sep 09 2018: (Start)
(End)
Comments