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A356014
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Consider the exponents in the prime factorization of n, and replace each run of k consecutive e's by a unique e; the resulting list corresponds to the exponents in the prime factorization of a(n).
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3
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1, 2, 3, 4, 3, 2, 3, 8, 9, 10, 3, 12, 3, 10, 3, 16, 3, 18, 3, 20, 21, 10, 3, 24, 9, 10, 27, 20, 3, 2, 3, 32, 21, 10, 3, 4, 3, 10, 21, 40, 3, 10, 3, 20, 45, 10, 3, 48, 9, 50, 21, 20, 3, 54, 21, 40, 21, 10, 3, 12, 3, 10, 63, 64, 21, 10, 3, 20, 21, 10, 3, 72, 3
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OFFSET
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1,2
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COMMENTS
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We ignore the exponents (all 0's) for the prime numbers beyond the greatest prime factor of n.
This sequence operates on prime exponents as A090079 and A337864 operate on binary and decimal digits, respectively.
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LINKS
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FORMULA
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a(a(n)) = a(n).
a(n^k) = a(n)^k for any k >= 0.
a(n) = 3 iff n belongs to A294674 \ {1}.
a(n) = 4 iff n belongs to A061742 \ {1}.
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EXAMPLE
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For n = 99:
- 99 = 11^1 * 7^0 * 5^0 * 3^2 * 2^0,
- the list of exponents is: 1 0 0 2 0,
- compressing consecutive values, we obtain: 1 0 2 0,
- so a(99) = 7^1 * 5^0 * 3^2 * 2^0 = 63.
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PROG
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(PARI) a(n) = { my (v=1, e=-1, k=0); forprime (p=2, oo, if (n==1, return (v), if (e!=e=valuation(n, p), v*=prime(k++)^e); n/=p^e)) }
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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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