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 A370180 Size of the group Z_3*/(Z_3*)^n, where Z_3 is the ring of 3-adic integers. 5
 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 (list; graph; refs; listen; history; text; internal format)
 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

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Last modified August 11 09:55 EDT 2024. Contains 375059 sequences. (Running on oeis4.)