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A086435
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Maximum number of parts possible in a factorization of n into a product of distinct numbers > 1.
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4
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0, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 2, 3, 1, 2, 2, 3, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 2, 3, 2, 2, 1, 3, 1, 2, 2, 3, 2, 3, 1, 2, 2, 3, 1, 3, 1, 2, 2, 2, 2, 3, 1, 3, 2, 2, 1, 3, 2, 2, 2, 3, 1, 3, 2, 2, 2, 2, 2, 3, 1, 2, 2, 3, 1, 3, 1, 3, 3
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,6
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COMMENTS
| For n>1, a((n+1)!) = n is the first occurrence of n in the sequence. This function depends only on the prime signature of n. - Franklin T. Adams-Watters (FrankTAW(AT)Netscape.net), Dec 19 2006
For integer n and prime p not dividing n, a(n*p) = a(n) + 1. [From Max Alekseyev (maxale(AT)gmail.com), Apr 23 2010]
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LINKS
| Eric Weisstein's World of Mathematics, UnorderedFactorization
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EXAMPLE
| a(6)=2 since 6 may be factored into distinct parts as {{2,3},{6}}, so the largest number of factors possible is 2.
a(8)=2 since 8 may be factored into distinct parts as {{8},{2,4}}, so the largest numbers of factors possible is 2.
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PROG
| (PARI) { a(n, m=1) = if(n>m, 1 + vecmax( apply( x->if(x>m, a(n/x, x)), divisors(n) ))) } [From Max Alekseyev (maxale(AT)gmail.com), Jul 16 2009]
(PARI) { aopt(n) = local(f, t); f=factor(n)[, 2]; t=select(x->x>1, f); a(prod(j=1, #t, prime(j)^t[j])) + #f - #t } /* optimized version */ [From Max Alekseyev (maxale(AT)gmail.com), Apr 23 2010]
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CROSSREFS
| Cf. A000142, A025487.
Sequence in context: A065031 A058061 A064547 * A099305 A033109 A175096
Adjacent sequences: A086432 A086433 A086434 * A086436 A086437 A086438
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KEYWORD
| nonn
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AUTHOR
| Eric Weisstein (eric(AT)weisstein.com), Jul 19, 2003
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EXTENSIONS
| More terms from Max Alekseyev (maxale(AT)gmail.com), Apr 23 2010
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