A160595 Numerator of resilience R(n) = phi(n)/(n-1), with a(1) = 1 by convention.

1, 1, 1, 2, 1, 2, 1, 4, 3, 4, 1, 4, 1, 6, 4, 8, 1, 6, 1, 8, 3, 10, 1, 8, 5, 12, 9, 4, 1, 8, 1, 16, 5, 16, 12, 12, 1, 18, 12, 16, 1, 12, 1, 20, 6, 22, 1, 16, 7, 20, 16, 8, 1, 18, 20, 24, 9, 28, 1, 16, 1, 30, 18, 32, 3, 4, 1, 32, 11, 8, 1, 24, 1, 36, 20, 12, 15, 24, 1, 32, 27, 40, 1, 24, 16, 42, 28

Offset: 1

Author

M. F. Hasler, May 23 2009

Keywords

• nonn
• look

Comments

The resilience of a denominator, R(d), is the ratio of proper fractions n/d, 0 < n < d, that are resilient, i.e., such that gcd(n, d) = 1. Obviously this is the case for phi(d) proper fractions among the d - 1 possible ones.

a(n) = 1 if n is prime. It is unknown whether there exist composite n with a(n) = 1 (see Wikipedia link). - Robert Israel, Dec 26 2016

Conjecture: a(n) > 2 for every composite n > 6. Slightly stronger than the Lehmer's totient conjecture (1932). - Thomas Ordowski, Mar 13 2019

Examples

			a(9) = 3 since for the denominator d = 9, among the 8 proper fractions n/9 (n = 1, ..., 8), six cannot be canceled down by a common factor (namely 1/9, 2/9, 4/9, 5/9, 7/9, 8/9), thus R(9) = 6/8 = 3/4.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537 (terms 2..10000 from Robert Israel)
• Project Euler, Problem 245: resilient fractions, May 2009
• Wikipedia, Lehmer's totient problem.

Crossrefs

Cf. A000010, A049559, A247074, A318829.

Programs

Magma

[Numerator(EulerPhi(n)/(n-1)): n in [2..90]]; // Vincenzo Librandi, Jan 02 2017

Maple

seq(numer(numtheory:-phi(n)/(n-1)),n=2..100); # Robert Israel, Dec 26 2016

Mathematica

Numerator[Table[EulerPhi[n]/(n - 1), {n, 2, 87}]] (* Alonso del Arte, Sep 19 2011 *)

PARI

A160595(n) = if(1==n,n,numerator(eulerphi(n)/(n-1)));

Formula

a(n) = phi(n)/gcd(phi(n),n-1) = A000010(n) / A049559(n) = A247074(n) * A318829(n). - Antti Karttunen, Sep 09 2018

Extensions

Term a(1) = 1 prepended by Antti Karttunen, Sep 09 2018