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

1,2

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

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)

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 *)

Prog

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

Crossrefs

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

Cf. A370180.

Sequence in context: A134902 A110992 A199203 * A370567 A371500 A197580

Adjacent sequences: A370562 A370563 A370564 * A370566 A370567 A370568

Keyword

nonn,easy,mult

Author

Jianing Song, Apr 30 2024

Status

approved