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

1,2

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.

Links

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

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)

Example

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.

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 *)

Prog

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

Crossrefs

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

Cf. A370567.

Sequence in context: A355624 A065369 A336404 * A167772 A077870 A294408

Adjacent sequences: A370179 A370180 A370181 * A370183 A370184 A370185

Keyword

nonn,easy,mult

Author

Jianing Song, Apr 30 2024

Status

approved