OFFSET
1,2
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
FORMULA
a(n) = A025473(n + 1) - 1. - Bill McEachen, Sep 11 2021
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 *)
PROG
(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
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Labos Elemer, Aug 27 2001
STATUS
approved