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

1,1

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

Table of n, a(n) for n=1..11.

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.

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]]]

Crossrefs

Cf. A001222, A067665, A068796.

Sequence in context: A343806 A372254 A210654 * A254639 A049951 A025263

Adjacent sequences: A068794 A068795 A068796 * A068798 A068799 A068800

Keyword

more,nonn

Author

Dean Hickerson, Mar 05 2002

Extensions

a(11) from Donovan Johnson, Oct 15 2008

Status

approved