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A056169
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Number of unitary prime divisors of n.
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26
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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; refs; listen; history; internal format)
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OFFSET
| 1,6
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COMMENTS
| 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
| T. D. Noe, Table of n, a(n) for n=1..10000
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FORMULA
| 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
| Cf. A001694, A076445, A002110, A034444, A001221.
Sequence in context: A125676 A025874 A050326 * A125070 A125071 A177207
Adjacent sequences: A056166 A056167 A056168 * A056170 A056171 A056172
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KEYWORD
| nice,nonn
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AUTHOR
| Labos E. (labos(AT)ana.sote.hu), Jul 27 2000
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