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Search: id:A063872
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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
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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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
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EXAMPLE
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For primes quotient=(p-1)/1 = p-1; for p^a,a>1:quotient=p^(a-1)[p-1]/p^(a-1)=p-1 So p-1 values occur repeatedly: Seq+1=Primes.
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CROSSREFS
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Cf. A000010, A051953, A007694, A000961, A054740, A049237.
Sequence in context: A137424 A083007 A002987 this_sequence A033884 A062344 A033877
Adjacent sequences: A063869 A063870 A063871 this_sequence A063873 A063874 A063875
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KEYWORD
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easy,nonn
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AUTHOR
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Labos E. (labos(AT)ana.sote.hu), Aug 27 2001
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