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A365490
The number of divisors of the largest 4th power dividing n.
1
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 5, 1, 1, 1, 1, 1, 1
OFFSET
1,16
COMMENTS
The number of divisors of the 4th root of the largest 4th power dividing n, A053164(n), is A063775(n).
LINKS
FORMULA
a(n) = A000005(A008835(n)).
Multiplicative with a(p^e) = 4*floor(e/4) + 1.
a(n) = 1 if and only if n is a biquadratefree number (A046100).
a(n) <= A000005(n) with equality if and only if n is a fourth power (A000583).
Dirichlet g.f.: zeta(s) * zeta(4*s) * Product_{p prime} (1 + 3/p^(4*s)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = zeta(4) * Product_{p prime} (1 + 3/p^4) = 1.3414590511076... . In general, the asymptotic mean of the number of divisors of the largest k-th power dividing n is zeta(k) * Product_{p prime} (1 + (k-1)/p^k).
MATHEMATICA
f[p_, e_] := 4*Floor[e/4] + 1; a[n_] := Times @@ f @@@ FactorInteger[n]; a[1] = 1; Array[a, 100]
PROG
(PARI) a(n) = vecprod(apply(x -> 4*(x\4) + 1, factor(n)[, 2]));
(Python)
from math import prod
from sympy import factorint
def A365490(n): return prod(e&-4|1 for e in factorint(n).values()) # Chai Wah Wu, Aug 08 2024
CROSSREFS
KEYWORD
nonn,easy,mult
AUTHOR
Amiram Eldar, Sep 05 2023
STATUS
approved