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A068797
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Minimum x such that f(x)=n, where f(x)=A068796(x) is the maximum k such that k consecutive integers starting at x have distinct numbers of prime factors (counted with multiplicity).
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2
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2, 1, 6, 15, 60, 726, 6318, 189375, 755968, 683441871, 33714015615
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| The number of prime factors (counted with multiplicity) of n is bigomega(n) = A001222(n).
The known terms, except for the first, agree with A067665. Is that true forever?
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REFERENCES
| J.-M. De Koninck, J. B. Friedlander, and F. Luca, On strings of consecutive integers with a distinct number of prime factors, Proc. Amer. Math. Soc., 137 (2009), 1585-1592.
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MATHEMATICA
| 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]]]; a[n_] := For[x=1, True, x++, If[f[x]==n, Return[x]]]
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CROSSREFS
| Cf. A001222, A067665, A068796.
Sequence in context: A002562 A136456 A123968 * A049951 A025263 A097947
Adjacent sequences: A068794 A068795 A068796 * A068798 A068799 A068800
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KEYWORD
| more,nonn
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AUTHOR
| Dean Hickerson (dean.hickerson(AT)yahoo.com), Mar 05 2002
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EXTENSIONS
| a(11) from Donovan Johnson (donovan.johnson(AT)yahoo.com), Oct 15 2008
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