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The number of infinitary divisors of n that are cubefree.
1

%I #17 Jan 09 2024 08:46:51

%S 1,2,2,2,2,4,2,3,2,4,2,4,2,4,4,1,2,4,2,4,4,4,2,6,2,4,3,4,2,8,2,2,4,4,

%T 4,4,2,4,4,6,2,8,2,4,4,4,2,2,2,4,4,4,2,6,4,6,4,4,2,8,2,4,4,2,4,8,2,4,

%U 4,8,2,6,2,4,4,4,4,8,2,2,1,4,2,8,4,4,4

%N The number of infinitary divisors of n that are cubefree.

%C The number of infinitary divisors of n that are squarefree is A055076(n).

%H Amiram Eldar, <a href="/A368883/b368883.txt">Table of n, a(n) for n = 1..10000</a>

%H Vaclav Kotesovec, <a href="/A368883/a368883.jpg">Graph - the asymptotic ratio (100000 terms)</a>

%F Multiplicative with a(p^e) = 2 if e == 1 or 2 (mod 4), 3 if e == 3 (mod 4), and 1 if e == 0 (mod 4).

%F a(n) >= 1, with equality if and only if n is a 4th power (A000583).

%F a(n) <= A037445(n), with equality if and only if n is cubefree (A004709).

%F Dirichlet g.f.: zeta(4*s) * Product_{p prime} (1 + 2/p^s + 2/p^(2*s) + 3/p^(3*s)).

%F From _Vaclav Kotesovec_, Jan 09 2024: (Start)

%F Dirichlet g.f.: zeta(4*s) * zeta(s)^2 * Product_{p prime} (1 - 1/p^(2*s) + 1/p^(3*s) - 4/p^(4*s) + 3/p^(5*s)).

%F Let f(s) = Product_{p prime} (1 - 1/p^(2*s) + 1/p^(3*s) - 4/p^(4*s) + 3/p^(5*s)).

%F Sum_{k=1..n} a(k) ~ f(1) * zeta(4) * n * (log(n) + 2*gamma - 1 + f'(1)/f(1) + 4*zeta'(4)/zeta(4)), where

%F f(1) = Product_{p prime} (1 - 1/p^2 + 1/p^3 - 4/p^4 + 3/p^5) = 0.5857770602270641007515615375942370402509903724261557972367075945186871...,

%F f'(1) = f(1) * Sum_{p prime} (2*p^2 - p + 15) * log(p) / (p^4 + p^3 + p - 3) = f(1) * 1.319786264712492218167871116508220489817987315752197198819256094...,

%F gamma is the Euler-Mascheroni constant A001620, zeta(4) = Pi^4/90 = A013662 and for zeta'(4) see A261506. (End)

%t f[p_, e_] := Switch[Mod[e, 4], 1, 2, 2, 2, 3, 3, 0, 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

%o (PARI) a(n) = vecprod(apply(x -> [1,2,2,3][x%4+1], factor(n)[, 2]));

%Y Cf. A000583, A004709, A037445, A055076, A077609.

%K nonn,easy,mult

%O 1,2

%A _Amiram Eldar_, Jan 09 2024