A068796 Maximum k such that k consecutive integers starting at n have distinct numbers of prime factors (counted with multiplicity).

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

Offset: 1

Author

Dean Hickerson, Mar 05 2002

Keywords

• nonn

Comments

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

Examples

			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.
		

Links

• Zak Seidov, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A001222, A067665, A068797.

Programs

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]]]
ss={}; Do[s={PrimeOmega[n]};k=1;While[FreeQ[s, (b=PrimeOmega[n+k])],s=AppendTo[s,b];k++];ss=AppendTo[ss,k],{n,103}]; (* Zak Seidov, Nov 09 2015 *)