cp's OEIS Frontend

A365490 The number of divisors of the largest 4th power dividing n.

%I A365490 #12 Aug 08 2024 14:28:08
%S A365490 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,
%T A365490 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,
%U A365490 1,1,1,1,1,1,1,1,1,1,1,5,5,1,1,1,1,1,1
%N A365490 The number of divisors of the largest 4th power dividing n.
%C A365490 The number of divisors of the 4th root of the largest 4th power dividing n, A053164(n), is A063775(n).
%H A365490 Amiram Eldar, <a href="/A365490/b365490.txt">Table of n, a(n) for n = 1..10000</a>
%F A365490 a(n) = A000005(A008835(n)).
%F A365490 Multiplicative with a(p^e) = 4*floor(e/4) + 1.
%F A365490 a(n) = 1 if and only if n is a biquadratefree number (A046100).
%F A365490 a(n) <= A000005(n) with equality if and only if n is a fourth power (A000583).
%F A365490 Dirichlet g.f.: zeta(s) * zeta(4*s) * Product_{p prime} (1 + 3/p^(4*s)).
%F A365490 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).
%t A365490 f[p_, e_] := 4*Floor[e/4] + 1; a[n_] := Times @@ f @@@ FactorInteger[n]; a[1] = 1; Array[a, 100]
%o A365490 (PARI) a(n) = vecprod(apply(x -> 4*(x\4) + 1, factor(n)[, 2]));
%o A365490 (Python)
%o A365490 from math import prod
%o A365490 from sympy import factorint
%o A365490 def A365490(n): return prod(e&-4|1 for e in factorint(n).values()) # _Chai Wah Wu_, Aug 08 2024
%Y A365490 Cf. A000005, A000583, A046100, A008835, A053164, A063775.
%K A365490 nonn,easy,mult
%O A365490 1,16
%A A365490 _Amiram Eldar_, Sep 05 2023