A370565 Size of the group Q_3*/(Q_3*)^n, where Q_3 is the field of 3-adic numbers.

1, 4, 9, 8, 5, 36, 7, 16, 81, 20, 11, 72, 13, 28, 45, 32, 17, 324, 19, 40, 63, 44, 23, 144, 25, 52, 729, 56, 29, 180, 31, 64, 99, 68, 35, 648, 37, 76, 117, 80, 41, 252, 43, 88, 405, 92, 47, 288, 49, 100, 153, 104, 53, 2916, 55, 112, 171, 116, 59, 360, 61, 124, 567, 128

Offset: 1

Author

Jianing Song, Apr 30 2024

Keywords

• nonn
• easy
• mult

Comments

We have Q_3* = 3^Z X Z_3*, so Q_3*/(Q_3*)^k = (3^Z/3^(kZ)) X (Z_p*/(Z_3*)^k). Note that 3^Z/3^(kZ) is a cyclic group of order k. For the group structure of (Z_3*/(Z_3*)^k), see A370050.

Links

• Jianing Song, Table of n, a(n) for n = 1..10000

Crossrefs

Row 2 of A370067. Cf. A001620, A370050, A370564, A370566, A370567.

Cf. A370180.

Programs

Mathematica

a[n_] := Module[{e2 = IntegerExponent[n, 2], e3 = IntegerExponent[n, 3]}, 2^Min[e2, 1] * 3^e3 * n]; Array[a, 100] (* Amiram Eldar, May 20 2024 *)

PARI

a(n, {p=3}) = my(e = valuation(n, p)); n * p^e*gcd(p-1, n/p^e)

Formula

Write n = 3^e * n' with k' not being divisible by 3, then a(n) = n * 3^e * gcd(2,n').

Multiplicative with a(3^e) = 3^(2*e), a(2^e) = 2^(e+1) and a(p^e) = p^e for primes p != 2, 3.

a(n) = n * A370180(n).

From Amiram Eldar, May 20 2024: (Start)

Dirichlet g.f.: ((1 + 1/2^(s-1)) * (1 - 1/3^(s-1))/(1 - 1/3^(s-2))) * zeta(s-1).

Sum_{k=1..n} a(k) ~ (n^2/(2*log(3))) * (log(n) + gamma - 1/2 + log(3) - log(2)/3), where gamma is Euler's constant (A001620). (End)