A063872 Let m be the n-th positive integer such that phi(m) is divisible by m - phi(m). Then a(n) = phi(m)/(m - phi(m)).

1, 2, 1, 4, 6, 1, 2, 10, 12, 1, 16, 18, 22, 4, 2, 28, 30, 1, 36, 40, 42, 46, 6, 52, 58, 60, 1, 66, 70, 72, 78, 2, 82, 88, 96, 100, 102, 106, 108, 112, 10, 4, 126, 1, 130, 136, 138, 148, 150, 156, 162, 166, 12, 172, 178, 180, 190, 192, 196, 198, 210, 222, 226, 228, 232

Offset: 1

Author

Labos Elemer, Aug 27 2001

Keywords

• easy
• nonn

Comments

m is the n-th prime power larger than 1; i.e., m = A000961(n+1). Proof: If phi(m) is divisible by m-phi(m), then m is divisible by m-phi(m). Let k be the product of the distinct prime factors of m. Then phi(m)/m = phi(k)/k, so k/(k-phi(k)) = m/(m-phi(m)) is an integer. Thus k is divisible by k-phi(k) and k is squarefree. Let k-phi(k) = d and k/(k-phi(k)) = e; note that e>1 and GCD(d,e)=1. Thus d = k - phi(k) = d e - phi(d e) = d e - phi(d) phi(e) so d (e-1) = d e - d = phi(d) phi(e) <= phi(d) (e-1) and d <= phi(d). But this implies that d=1, so phi(k)=k-1 and k is prime. Hence m is a prime power. - Dean Hickerson, Aug 28 2001

For primes, quotient = (p - 1) / 1 = p - 1; for prime powers, p^a, a > 1: quotient = p^(a - 1)(p - 1) / p^(a - 1) = p - 1, so each p - 1 values occur infinitely often: a(n) + 1 = root of n-th prime power with positive exponent, i.e., A025473(n+1). - [Edited by] Daniel Forgues, May 08 2014

"LCM numeral system": a(n+1) is maximum digit for index n, n >= 0; a(-n) is maximum digit for index n, n < 0. - Daniel Forgues, May 03 2014

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000
• OEIS Wiki, LCM numeral system

Crossrefs

Cf. A000010, A051953, A007694, A000961, A054740, A049237, A025473.

Programs

Mathematica

epd[n_]:=Module[{ep=EulerPhi[n]},If[Divisible[ep,n-ep],ep/(n-ep),Nothing]]; Array[epd,300,2] (* Harvey P. Dale, Dec 27 2020 *)

PARI

M(n) = ispower(n, , &n); if (isprime(n), n, 1); \\ A014963
apply(x->x-1, select(isprime, apply(x->M(x+1), [1..260]))) \\ Michel Marcus, Sep 14 2021

Formula

a(n) = A025473(n + 1) - 1. - Bill McEachen, Sep 11 2021