A366765 The largest divisor of n that have no exponent 2 in their prime factorization.

1, 2, 3, 2, 5, 6, 7, 8, 3, 10, 11, 6, 13, 14, 15, 16, 17, 6, 19, 10, 21, 22, 23, 24, 5, 26, 27, 14, 29, 30, 31, 32, 33, 34, 35, 6, 37, 38, 39, 40, 41, 42, 43, 22, 15, 46, 47, 48, 7, 10, 51, 26, 53, 54, 55, 56, 57, 58, 59, 30, 61, 62, 21, 64, 65, 66, 67, 34, 69

Offset: 1

Author

Amiram Eldar, Oct 21 2023

Keywords

• nonn
• easy
• mult

Comments

The largest term of A337050 that divides n.

The number of these divisors is A366763(n), and their sum is A366764(n).

Links

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

Crossrefs

Cf. A337050, A366763, A366764.

Similar sequences: A055231, A057521, A008833, A350390.

Programs

Mathematica

f[p_, e_] := p^If[e < 3, 1, e]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

PARI

a(n) = {my(f = factor(n)); prod(i = 1, #f~, f[i, 1] ^ if(f[i, 2] < 3, 1, f[i, 2]));}

Formula

Multiplicative with a(p^e) = p if e <= 2 and p^e otherwise.

a(n) <= n, with equality if and only if n is in A337050.

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

Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = Product_{p prime} (1 - 1/p^2 + 2/p^3 - 1/p^4) = 0.83234421330425224469... .