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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
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OFFSET
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1,1
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COMMENTS
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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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LINKS
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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]]]; a[n_] := For[x=1, True, x++, If[f[x]==n, Return[x]]]
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CROSSREFS
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KEYWORD
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more,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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