A370182 Size of the group Z_7*/(Z_7*)^n, where Z_7 is the ring of 7-adic integers.

1, 2, 3, 2, 1, 6, 7, 2, 3, 2, 1, 6, 1, 14, 3, 2, 1, 6, 1, 2, 21, 2, 1, 6, 1, 2, 3, 14, 1, 6, 1, 2, 3, 2, 7, 6, 1, 2, 3, 2, 1, 42, 1, 2, 3, 2, 1, 6, 49, 2, 3, 2, 1, 6, 1, 14, 3, 2, 1, 6, 1, 2, 21, 2, 1, 6, 1, 2, 3, 14, 1, 6, 1, 2, 3, 2, 7, 6, 1, 2, 3, 2, 1, 42, 1, 2, 3, 2, 1, 6

Offset: 1

Author

Jianing Song, Apr 30 2024

Keywords

• nonn
• easy
• mult

Comments

We have that Z_7*/(Z_7*)^n is the inverse limit of (Z/7^iZ)*/((Z/7^iZ)*)^n as i tends to infinity. Write n = 7^e * n' with n' not being divisible by 7, then the group is cyclic of order 7^e * gcd(7,n'). See A370050.

Examples

			We have Z_7*/(Z_7*)^7 = Z_7* / ((1+49Z_7) U (18+49Z_7) U (19+49Z_7) U (30+49Z_7) U (31+49Z_7) U (48+49Z_7)) = (Z/49Z)*/((1+49Z) U (18+49Z) U (19+49Z) U (30+49Z) U (31+49Z) U (48+49Z)) = C_7, so a(7) = 7.
We have Z_7*/(Z_7*)^14 = Z_7* / ((1+49Z_7) U (18+49Z_7) U (30+49Z_7)) = (Z/49Z)*/((1+49Z) U (18+49Z) U (30+49Z)) = C_14, so a(14) = 14.
		

Links

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

Crossrefs

Row 4 of A370050. Cf. A001620, A297402, A370180, A370181.

Cf. A370567.

Programs

Mathematica

a[n_] := Module[{e2 = IntegerExponent[n, 2], e3 = IntegerExponent[n, 3], e7 = IntegerExponent[n, 7]}, 2^Min[e2, 1] * 3^Min[e3, 1] * 7^e7]; Array[a, 100] (* Amiram Eldar, May 20 2024 *)

PARI

a(n,{p=7}) = my(e = valuation(n, p)); p^e*gcd(p-1, n/p^e)

Formula

Multiplicative with a(7^e) = 7^e, a(2^e) = 2, a(3^e) = 3 and a(p^e) = 1 for primes p != 2, 3, 7.

From Amiram Eldar, May 20 2024: (Start)

Dirichlet g.f.: (1 + 1/2^s) * (1 + 2/3^s) * ((1 - 1/7^s)/(1 - 1/7^(s-1))) * zeta(s).

Sum_{k=1..n} a(k) ~ (15*n/(7*log(7))) * (log(n) + gamma - 1 + 2*log(7)/3 - 2*log(3)/5 - log(2)/3), where gamma is Euler's constant (A001620). (End)