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I think you compute A[n] and sumA[n] same timeCho wrote: 3.5 Compute all A[n] by making use of factor[n], O(n)
Cho wrote: >> p=s-sum(p[j]| i mod j=0, 0<j<i);
Is this part O(n sqrt(n))?
it depent from implimentation
In my implement not.
cose count action dec(p[j*i],p) i=1,2,3..mn and j=2i,3i,..flor(m/i)j
get for c(100000)=1066750
c(n)=sum(n div i -1, where 0<i<n)
equal around H(mn)*mn times
where H(n)=sum(1/i where 0<i<n)
so that part: >> p=s-sum(p[j]| i mod j=0, 0<j<i);
I think not O(n*sqrt(n)) but O(n*ln(n))