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Search: id:A068797
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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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+0
3
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| 2, 1, 6, 15, 60, 726, 6318, 189375, 755968, 683441871
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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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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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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Cf. A001222, A067665, A068796.
Sequence in context: A002562 A136456 A123968 this_sequence A049951 A025263 A097947
Adjacent sequences: A068794 A068795 A068796 this_sequence A068798 A068799 A068800
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KEYWORD
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more,nonn
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AUTHOR
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Dean Hickerson (dean(AT)math.ucdavis.edu), Mar 05 2002
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