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A332334
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Let a(1) = a(2) = 1, and for n > 2 let a(n) = p where p is the largest prime such that p# divides phi(n), where phi is Euler's totient function and # is the primorial.
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0
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1, 1, 2, 2, 2, 2, 3, 2, 3, 2, 2, 2, 3, 3, 2, 2, 2, 3, 3, 2, 3, 2, 2, 2, 2, 3, 3, 3, 2, 2, 5, 2, 2, 2, 3, 3, 3, 3, 3, 2, 2, 3, 3, 2, 3, 2, 2, 2, 3, 2, 2, 3, 2, 3, 2, 3, 3, 2, 2, 2, 5, 5, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 3, 3, 2, 3, 5, 3, 3, 2, 3, 2, 2, 3, 2, 3, 2
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OFFSET
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1,3
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COMMENTS
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Pollack and Pomerance show that the normal order of a(n) is log log n/log log log n. The maximal order is log n (for primorial primes A018239, by the prime number theorem) and the minimal order, for n > 2, is 2 (for products of Fermat primes A143512, apart from 1).
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LINKS
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PROG
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(PARI) a(n)=my(ph=eulerphi(n)); my(p=1); forprime(q=2, , if(ph%q, return(p), p=q))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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