A362852 The number of divisors of n that are both bi-unitary and exponential.
1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1
Offset: 1
Examples
a(8) = 2 since 8 has 2 divisors that are both bi-unitary and exponential: 2 and 8.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
- Eric Weisstein's World of Mathematics, Biunitary Divisor.
- Eric Weisstein's World of Mathematics, e-Divisor.
Crossrefs
Programs
-
Mathematica
f[p_, e_] := DivisorSigma[0, e] - If[OddQ[e], 0, 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
-
PARI
a(n) = {my(f = factor(n)); prod(i = 1, #f~, numdiv(f[i, 2]) - !(f[i, 2] % 2));}
Formula
Multiplicative with a(p^e) = d(e) if e is odd, and d(e)-1 if e is even, where d(k) is the number of divisors of k (A000005).
a(n) = 1 if and only if n is cubefree (A004709).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Product_{p prime} (1 + Sum_{k>=1} (d(k)+(k mod 2)-1)/p^k) = 1.1951330849... .
Comments