A380160 a(n) is the value of the Euler totient function when applied to the powerful part of n.
1, 1, 1, 2, 1, 1, 1, 4, 6, 1, 1, 2, 1, 1, 1, 8, 1, 6, 1, 2, 1, 1, 1, 4, 20, 1, 18, 2, 1, 1, 1, 16, 1, 1, 1, 12, 1, 1, 1, 4, 1, 1, 1, 2, 6, 1, 1, 8, 42, 20, 1, 2, 1, 18, 1, 4, 1, 1, 1, 2, 1, 1, 6, 32, 1, 1, 1, 2, 1, 1, 1, 24, 1, 1, 20, 2, 1, 1, 1, 8, 54, 1, 1, 2
Offset: 1
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
-
Mathematica
f[p_, e_] := If[e == 1, 1, (p-1)*p^(e-1)]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
-
PARI
a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i, 2] == 1, 1, (f[i, 1]-1)*f[i, 1]^(f[i, 2]-1)));}
Formula
a(n) >= 1, with equality if and only if n is squarefree (A005117).
Multiplicative with a(p) = 1, and a(p^e) = (p-1)*p^(e-1) if e >= 2.
Dirichlet g.f.: zeta(s-1) * Product_{p prime} (1 + 1/p^s - 1/p^(s-1) + 1/p^(2*s-2) - 2/p^(2*s-1)).
Sum_{k=1..n} a(k) ~ c * n^(3/2) / 3, where c = Product_{p prime} (1 + 2/p^(3/2) - 1/p^2 - 2/p^(5/2)) = 1.96428740396979919886... .