A369757 The number of divisors of the smallest cubefull exponentially odd number that is divisible by n.
1, 4, 4, 4, 4, 16, 4, 4, 4, 16, 4, 16, 4, 16, 16, 6, 4, 16, 4, 16, 16, 16, 4, 16, 4, 16, 4, 16, 4, 64, 4, 6, 16, 16, 16, 16, 4, 16, 16, 16, 4, 64, 4, 16, 16, 16, 4, 24, 4, 16, 16, 16, 4, 16, 16, 16, 16, 16, 4, 64, 4, 16, 16, 8, 16, 64, 4, 16, 16, 64, 4, 16, 4
Offset: 1
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Programs
-
Mathematica
f[p_, e_] := If[OddQ[e], Max[e, 3] + 1, e + 2]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
-
PARI
a(n) = vecprod(apply(x -> if(x%2, max(x, 3) + 1, x + 2), factor(n)[, 2]));
Formula
Multiplicative with a(p^e) = max(e,3) + 1 if e is odd, and e+2 if e is even.
Dirichlet g.f.: zeta(s) * zeta(2*s) * Product_{p prime} (1 + 3/p^s - 1/p^(2*s) - 3/p^(3*s) + 2/p^(4*s)).
From Vaclav Kotesovec, Feb 02 2024: (Start)
Dirichlet g.f.: zeta(s)^4 * Product_{p prime} (1 + (7*p^(2*s) + 2*p^(3*s) - 6*p^(4*s) - 7*p^s + 2) / ((p^s+1)*p^(5*s))).
Sum_{k=1..n} a(k) = c * n*log(n)^3/6 + O(n*log(n)^2), where c = Product_{p prime} (1 - (6*p^4 - 2*p^3 - 7*p^2 + 7*p - 2) / ((p+1)*p^5)) = 0.124604542136592401049820049658828040278... (End)