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A063872
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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)).
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2
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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
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OFFSET
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1,2
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COMMENTS
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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
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LINKS
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FORMULA
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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