A068797 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).

2, 1, 6, 15, 60, 726, 6318, 189375, 755968, 683441871, 33714015615

Offset: 1

Author

Dean Hickerson, Mar 05 2002

Keywords

• more
• nonn

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?

Links

• 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.

Crossrefs

Cf. A001222, A067665, A068796.

Programs

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]]]

Extensions

a(11) from Donovan Johnson, Oct 15 2008