A068316 Run lengths of the Moebius function applied to A051270 (numbers with 5 distinct prime factors).
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, 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, 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
Offset: 1
Keywords
Examples
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. 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, ...
Programs
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Maple
runl := 1 : for n from 2 to 1000 do if numtheory[mobius](A051270(n)) = numtheory[mobius](A051270(n-1)) then runl := runl+1 ; else printf("%d,",runl) ; runl := 1; end if; end do: # R. J. Mathar, Oct 13 2019
Extensions
Corrected and extended by R. J. Mathar, Oct 13 2019
Comments