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Search: id:A056169
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| A056169 |
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Number of unitary prime divisors of n. |
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+0
25
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| 0, 1, 1, 0, 1, 2, 1, 0, 0, 2, 1, 1, 1, 2, 2, 0, 1, 1, 1, 1, 2, 2, 1, 1, 0, 2, 0, 1, 1, 3, 1, 0, 2, 2, 2, 0, 1, 2, 2, 1, 1, 3, 1, 1, 1, 2, 1, 1, 0, 1, 2, 1, 1, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 0, 2, 3, 1, 1, 2, 3, 1, 0, 1, 2, 1, 1, 2, 3, 1, 1, 0, 2, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 1, 1, 1, 0, 1, 3, 1, 1, 3
(list; graph; listen)
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OFFSET
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1,6
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COMMENT
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The zeros of this sequences are the powerful numbers (A001694). There are no arbitrarily long subsequences with a given upper bound; for example, every sequence of 4 values includes one divisible by 2 but not 4, so there are no more than 3 consecutive zeros. Similarly, there can be no more than 23 consecutive values with none divisible by both 2 and 3 but neither 4 nor 9 (so a(n) >= 2), etc. In general, this gives an upper bound that is a (relatively) small multiple of the k-th primorial number (prime(k)#). One suspects that the actual upper bounds for such subsequences are quite a bit lower; e.g., Erdos conjectured that there are no three consecutive powerful numbers. - Frank Adams-Watters (FrankTAW(AT)Netscape.net), Aug 08 2006
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LINKS
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T. D. Noe, Table of n, a(n) for n=1..10000
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FORMULA
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A prime factor of n is unitary iff its exponent is 1 in prime factorization of n. In general GCD[p, n/p]=1 or =p
Additive with a(p^e) = 1 if e = 1, 0 otherwise.
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CROSSREFS
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Cf. A001694, A076445, A002110, A034444, A001221.
Adjacent sequences: A056166 A056167 A056168 this_sequence A056170 A056171 A056172
Sequence in context: A125676 A025874 A050326 this_sequence A125070 A125071 A136176
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KEYWORD
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nice,nonn
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AUTHOR
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Labos E. (labos(AT)ana.sote.hu), Jul 27 2000
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