A384554 The sum of the infinitary divisors of n that are cubefree.
1, 3, 4, 5, 6, 12, 8, 7, 10, 18, 12, 20, 14, 24, 24, 1, 18, 30, 20, 30, 32, 36, 24, 28, 26, 42, 13, 40, 30, 72, 32, 3, 48, 54, 48, 50, 38, 60, 56, 42, 42, 96, 44, 60, 60, 72, 48, 4, 50, 78, 72, 70, 54, 39, 72, 56, 80, 90, 60, 120, 62, 96, 80, 5, 84, 144, 68, 90
Offset: 1
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Mathematica
f[p_, e_] := Switch[Mod[e, 4], 0, 1, 1, p+1, 2, p^2+1, 3, p^2+p+1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
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PARI
a(n) = {my(f = factor(n)); prod(i = 1, #f~, p = f[i,1]; e = f[i,2]; [1, p+1, p^2+1, p^2+p+1][e%4+1]);}
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Python
from math import prod from sympy import factorint def A384554(n): return prod((1,p+1,p**2+1,p*(p+1)+1)[e&3] for p,e in factorint(n).items()) # Chai Wah Wu, Jun 03 2025
Formula
Multiplicative with a(p^e) = 1 if e == 0 (mod 4), p + 1 if e == 1 (mod 4), p^2 + 1 if e == 2 (mod 4), and p^2 + p + 1 if e == 3 (mod 4).
Dirichlet g.f.: zeta(4*s) * Product_{p prime} (1 + 1/p^(s-1) + 1/p^s + 1/p^(2*s-2) + 1/p^(2*s) + 1/p^(3*s-1) + 1/p^(3*s-2) + 1/p^(3*s)).
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = zeta(8) * Product_{p prime} (1 + 1/p^2 - 2/p^3 + 2/p^4 - 1/p^5 - 1/p^7) = 1.2351002232125595782019... .
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