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A068316 Run lengths of the Moebius function applied to A051270 (numbers with 5 distinct prime factors).

%I A068316 #15 Oct 14 2019 01:29:26
%S A068316 5,1,1,1,6,2,4,3,4,1,2,1,6,2,1,1,1,1,1,2,1,1,2,1,2,2,2,1,3,2,2,1,2,1,
%T A068316 3,1,2,1,3,2,2,1,1,1,1,1,4,1,2,2,3,1,2,5,2,2,1,1,2,4,1,1,2,1,1,2,1,1,
%U A068316 3,1,2,2,4,1,2,2,2,1,4,2,2,1,2,2,2,1,2,1,2,1,1,1,3,1,1,1,1,2
%N A068316 Run lengths of the Moebius function applied to A051270 (numbers with 5 distinct prime factors).
%e A068316 If we consider A051270 and apply the Moebius function mu(n) to it we get a sequence of values: (-1,-1,-1,-1,-1),0,(-1),0,(-1,-1,-1,-1,-1,-1),0,0,(-1,-1,-1,-1),0,0,0,(-1,-1,-1,-1),0,(-1,-1),0,(-1, ... If we then look at the lengths of runs of equal terms, we get the sequence.
%e A068316 If we consider the values of A051270 which are not in A046387 we get numbers which are not squarefree, so mu(A051270(.)) is zero: 4620, 5460, 6930, ...
%p A068316 runl := 1 :
%p A068316 for n from 2 to 1000 do
%p A068316     if numtheory[mobius](A051270(n)) = numtheory[mobius](A051270(n-1)) then
%p A068316         runl := runl+1 ;
%p A068316     else
%p A068316         printf("%d,",runl) ;
%p A068316         runl := 1;
%p A068316     end if;
%p A068316 end do: # _R. J. Mathar_, Oct 13 2019
%Y A068316 Cf. A046387, A051270.
%K A068316 nonn
%O A068316 1,1
%A A068316 _Jani Melik_, Feb 26 2002
%E A068316 Corrected and extended by _R. J. Mathar_, Oct 13 2019