A333806 Number of distinct prime divisors of n that are < sqrt(n).
0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 2, 0, 1, 1, 1, 0, 2, 0, 1, 1, 1, 0, 2, 0, 1, 1, 1, 0, 3, 0, 1, 1, 1, 1, 2, 0, 1, 1, 2, 0, 2, 0, 1, 2, 1, 0, 2, 0, 2, 1, 1, 0, 2, 1, 2, 1, 1, 0, 3, 0, 1, 2, 1, 1, 2, 0, 1, 1, 3, 0, 2, 0, 1, 2, 1, 1, 2, 0, 2, 1, 1, 0, 3, 1, 1, 1, 1, 0, 3
Offset: 1
Keywords
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
Programs
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Maple
N:= 100: # for a(1)..a(N) V:= Vector(N): p:= 1: do p:= nextprime(p); if p^2 >= N then break fi; L:= [seq(p*k,k=p+1..N/p)]; V[L]:= V[L]+~1 od: convert(V,list); # Robert Israel, Apr 07 2020
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Mathematica
Table[DivisorSum[n, 1 &, # < Sqrt[n] && PrimeQ[#] &], {n, 1, 90}] nmax = 90; CoefficientList[Series[Sum[x^(Prime[k] (Prime[k] + 1))/(1 - x^Prime[k]), {k, 1, nmax}], {x, 0, nmax}], x] // Rest
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PARI
a(n)=my(f=factor(n)[,1]); sum(i=1,#f, f[i]^2
Charles R Greathouse IV, Apr 07 2020
Formula
G.f.: Sum_{k>=1} x^(prime(k)*(prime(k) + 1)) / (1 - x^prime(k)).
a(k*n) >= a(n) for k > 0. a(n^e) = A001221(n) for e > 2. - Richard Peterson, Dec 19 2024
Comments