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A068796 Maximum k such that k consecutive integers starting at n have distinct numbers of prime factors (counted with multiplicity). 2
2, 1, 2, 2, 2, 3, 3, 2, 1, 3, 2, 3, 2, 1, 4, 3, 2, 2, 3, 2, 1, 3, 3, 2, 1, 2, 1, 2, 2, 4, 3, 2, 1, 1, 3, 3, 2, 1, 4, 3, 2, 2, 2, 1, 4, 3, 4, 3, 2, 2, 4, 4, 3, 2, 2, 2, 1, 3, 2, 5, 4, 3, 3, 4, 3, 2, 3, 2, 4, 3, 2, 4, 3, 2, 1, 2, 5, 5, 4, 4, 3, 3, 3, 2, 1, 1, 3, 2, 4, 3, 2, 2, 1, 1, 4, 3, 2, 1, 3, 3, 2, 3, 4 (list; graph; refs; listen; history; internal format)
OFFSET

1,1

COMMENTS

The number of prime factors (counted with multiplicity) of n is bigomega(n) = A001222(n).

EXAMPLE

a(6)=3 because 6, 7, 8 and 9 have, respectively, 2, 1, 3 and 2 prime factors; the first 3 of these are distinct.

MATHEMATICA

bigomega[n_] := Plus@@Last/@FactorInteger[n]; a[n_] := For[k=1; s={bigomega[n]}, True, k++, If[MemberQ[s, z=bigomega[n+k]], Return[k], AppendTo[s, z]]]

CROSSREFS

Cf. A001222, A067665, A068797.

Sequence in context: A029232 A025807 A120254 * A154804 A053276 A064065

Adjacent sequences:  A068793 A068794 A068795 * A068797 A068798 A068799

KEYWORD

nonn

AUTHOR

Dean Hickerson (dean.hickerson(AT)yahoo.com), Mar 05 2002

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Last modified February 17 12:34 EST 2012. Contains 206018 sequences.