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Search: id:A067665
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| A067665 |
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The start of a record-setting run of consecutive integers with distinct numbers of prime factors (counted with multiplicity); i.e. let f(n) = A068796(n) be the maximum k such that k consecutive integers starting at n have distinct numbers of prime factors. Then n is in the sequence if f(n) is larger than f(m) for all m such that 1 <= m < n. |
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+0
5
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| 1, 6, 15, 60, 726, 6318, 189375, 755968, 683441871, 33714015615
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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The number of prime factors (counted with multiplicity) of n is bigomega(n) = A001222(n).
For the known terms, f(a(n)) = n+1. Is that true for all n?
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EXAMPLE
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The values of f(n) for n=1 to 15 are 2,1,2,2,2,3,3,2,1,3,2,3,2,1,4. Records occur at f(1)=2, f(6)=3 and f(15)=4.
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MATHEMATICA
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bigomega[n_] := Plus@@Last/@FactorInteger[n]; f[n_] := For[k=1; s={bigomega[n]}, True, k++, If[MemberQ[s, z=bigomega[n+k]], Return[k], AppendTo[s, z]]]; For[n=1; max=0, True, n++, If[f[n]>max, Print[n, " ", max=f[n]]]]
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CROSSREFS
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Cf. A001222, A067650, A068796, A068797.
Sequence in context: A073065 A093042 A145638 this_sequence A012595 A012294 A069750
Adjacent sequences: A067662 A067663 A067664 this_sequence A067666 A067667 A067668
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KEYWORD
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more,nonn
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AUTHOR
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G. L. Honaker, Jr. (honak3r(AT)gmail.com), Feb 03 2002
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EXTENSIONS
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More terms from Shyam Sunder Gupta (guptass(AT)rediffmail.com), Feb 08 2002
Edited by Robert G. Wilson v (rgwv(AT)rgwv.com), Feb 20 2002; and by Dean Hickerson (dean.hickerson(AT)yahoo.com), Mar 05 2002
a(10) from Donovan Johnson (donovan.johnson(AT)yahoo.com), Oct 15 2008
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