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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).
2

%I #14 Oct 25 2017 06:01:44

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

%N 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).

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

%C The known terms, except for the first, agree with A067665. Is that true forever?

%H J.-M. De Koninck, J. B. Friedlander, and F. Luca, <a href="http://dx.doi.org/10.1090/S0002-9939-08-09702-5">On strings of consecutive integers with a distinct number of prime factors</a>, Proc. Amer. Math. Soc., 137 (2009), 1585-1592.

%t 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]]]

%Y Cf. A001222, A067665, A068796.

%K more,nonn

%O 1,1

%A _Dean Hickerson_, Mar 05 2002

%E a(11) from _Donovan Johnson_, Oct 15 2008