A370180 Size of the group Z_3*/(Z_3*)^n, where Z_3 is the ring of 3-adic integers.

1, 2, 3, 2, 1, 6, 1, 2, 9, 2, 1, 6, 1, 2, 3, 2, 1, 18, 1, 2, 3, 2, 1, 6, 1, 2, 27, 2, 1, 6, 1, 2, 3, 2, 1, 18, 1, 2, 3, 2, 1, 6, 1, 2, 9, 2, 1, 6, 1, 2, 3, 2, 1, 54, 1, 2, 3, 2, 1, 6, 1, 2, 9, 2, 1, 6, 1, 2, 3, 2, 1, 18, 1, 2, 3, 2, 1, 6, 1, 2, 81, 2, 1, 6, 1, 2, 3, 2, 1, 18

Offset

1,2

Comments

We have that Z_3*/(Z_3*)^n is the inverse limit of (Z/3^iZ)*/((Z/3^iZ)*)^n as i tends to infinity. Write n = 3^e * n' with n' not being divisible by 3, then the group is cyclic of order 3^e * gcd(2,n'). See A370050.

Links

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

Formula

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

From Amiram Eldar, May 20 2024: (Start)

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

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

Example

We have Z_3*/(Z_3*)^3 = Z_3* / ((1+9Z_3) U (8+9Z_3)) = (Z/9Z)*/((1+9Z) U (8+9Z)) = C_3, so a(3) = 3.
We have Z_3*/(Z_3*)^6 = Z_3* / (1+9Z_3) = (Z/9Z)*/(1+9Z) = C_6, so a(6) = 6.

Mathematica

a[n_] := Module[{e2 = IntegerExponent[n, 2], e3 = IntegerExponent[n, 3]}, 2^If[e2 == 0, 0, 1] * 3^e3]; Array[a, 100] (* Amiram Eldar, May 20 2024 *)

Prog

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

Crossrefs

Row 2 of A370050. Cf. A001620, A297402, A370181, A370182.

Cf. A370565.

Sequence in context: A228549 A079893 A324646 * A286156 A113908 A355624

Adjacent sequences: A370177 A370178 A370179 * A370181 A370182 A370183

Keyword

nonn,easy,mult

Author

Jianing Song, Apr 30 2024

Status

approved