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A369716
The number of divisors of the smallest powerful number that is a multiple of n.
4
1, 3, 3, 3, 3, 9, 3, 4, 3, 9, 3, 9, 3, 9, 9, 5, 3, 9, 3, 9, 9, 9, 3, 12, 3, 9, 4, 9, 3, 27, 3, 6, 9, 9, 9, 9, 3, 9, 9, 12, 3, 27, 3, 9, 9, 9, 3, 15, 3, 9, 9, 9, 3, 12, 9, 12, 9, 9, 3, 27, 3, 9, 9, 7, 9, 27, 3, 9, 9, 27, 3, 12, 3, 9, 9, 9, 9, 27, 3, 15, 5, 9, 3
OFFSET
1,2
FORMULA
a(n) = A000005(A197863(n)).
Multiplicative with a(p) = 3 and a(p^e) = e+1 for e >= 2.
a(n) >= A000005(n), with equality if and only if n is powerful (A001694).
Dirichlet g.f.: zeta(s)^2 * Product_{p prime} (1 + 1/p^s - 2/p^(2*s) + 1/p^(3*s)).
From Vaclav Kotesovec, Jan 30 2024: (Start)
Dirichlet g.f.: zeta(s)^3 * Product_{p prime} (1 - 3/p^(2*s) + 3/p^(3*s) - 1/p^(4*s)).
Let f(s) = Product_{primes p} (1 - 3/p^(2*s) + 3/p^(3*s) - 1/p^(4*s)).
Sum_{k=1..n} a(k) ~ n * (f(1)*log(n)^2/2 + log(n)*((3*gamma - 1)*f(1) + f'(1)) + f(1)*(1 - 3*gamma + 3*gamma^2 - 3*sg1) + (3*gamma - 1)*f'(1) + f''(1)/2), where
f(1) = Product_{primes p} (1 - 3/p^2 + 3/p^3 - 1/p^4) = 0.33718787379158997196169281615215824494915412775816393888028828465611936...,
f'(1) = f(1) * Sum_{primes p} (6*p^2 - 9*p + 4) * log(p) / (p^4 - 3*p^2 + 3*p - 1) = f(1) * 2.35603132119230949914708478515883136510141335620960622673206366...,
f''(1) = f'(1)^2/f(1) + f(1) * Sum_{primes p} (-p*(12*p^5 - 27*p^4 + 16*p^3 + 9*p^2 - 12*p + 3) * log(p)^2 / (p^4 - 3*p^2 + 3*p - 1)^2) = f'(1)^2/f(1) + f(1) * (-7.3049026768735124341194605967271037971153161932236518820258070165876...),
gamma is the Euler-Mascheroni constant A001620 and sg1 is the first Stieltjes constant (see A082633). (End)
MATHEMATICA
f[p_, e_] := If[e == 1, 3, e + 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
PROG
(PARI) a(n) = vecprod(apply(x -> if(x == 1, 3, x+1), factor(n)[, 2]));
CROSSREFS
KEYWORD
nonn,easy,mult
AUTHOR
Amiram Eldar, Jan 30 2024
STATUS
approved