A368712 The maximal exponent in the prime factorization of the cubefree numbers.
0, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 2, 2, 1, 1
Offset: 1
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Mathematica
s[n_] := If[n == 1, 0, Max @@ Last /@ FactorInteger[n]]; s /@ Select[Range[100], Max[FactorInteger[#][[;; , 2]]] < 3 &] (* or *) f[n_] := Module[{e = Max @@ FactorInteger[n][[;; , 2]]}, If[e < 3, e, Nothing]]; f[1] = 0; Array[f, 100]
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PARI
lista(kmax) = {my(e); print1(0, ", "); for(k = 2, kmax, e = vecmax(factor(k)[,2]); if(e < 3, print1(e, ", ")));}
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Python
from sympy import mobius, integer_nthroot, factorint def A368712(n): def f(x): return n+x-sum(mobius(k)*(x//k**3) for k in range(1, integer_nthroot(x,3)[0]+1)) m, k = n, f(n) while m != k: m, k = k, f(k) return max(factorint(m).values(),default=0) # Chai Wah Wu, Aug 12 2024
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