A384555 The largest infinitary divisor of n that is cubefree.

1, 2, 3, 4, 5, 6, 7, 4, 9, 10, 11, 12, 13, 14, 15, 1, 17, 18, 19, 20, 21, 22, 23, 12, 25, 26, 9, 28, 29, 30, 31, 2, 33, 34, 35, 36, 37, 38, 39, 20, 41, 42, 43, 44, 45, 46, 47, 3, 49, 50, 51, 52, 53, 18, 55, 28, 57, 58, 59, 60, 61, 62, 63, 4, 65, 66, 67, 68, 69

Offset: 1

Author

Amiram Eldar, Jun 03 2025

Keywords

• nonn
• easy
• mult

Comments

The number of these divisors is A368883(n), and their sum is A384554(n).

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A004709, A007948, A013666, A077609, A368883, A384554.

Programs

Mathematica

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

PARI

a(n) = {my(f = factor(n)); prod(i = 1, #f~, p = f[i,1]; e = f[i,2]; [1, p, p^2, p^2][e%4+1]);}

Formula

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

a(n) = n if and only if n is cubefree (A004709).

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

Sum_{k=1..n} a(k) = c * n^2 / 2, where c = zeta(8) * Product_{p prime} (1 - 1/p^3 + 1/p^4 - 1/p^5) = 0.87406992849637563411... .