A370181 Size of the group Z_5*/(Z_5*)^n, where Z_5 is the ring of 5-adic integers.

1, 2, 1, 4, 5, 2, 1, 4, 1, 10, 1, 4, 1, 2, 5, 4, 1, 2, 1, 20, 1, 2, 1, 4, 25, 2, 1, 4, 1, 10, 1, 4, 1, 2, 5, 4, 1, 2, 1, 20, 1, 2, 1, 4, 5, 2, 1, 4, 1, 50, 1, 4, 1, 2, 5, 4, 1, 2, 1, 20, 1, 2, 1, 4, 5, 2, 1, 4, 1, 10, 1, 4, 1, 2, 25, 4, 1, 2, 1, 20, 1, 2, 1, 4, 5, 2, 1, 4, 1, 10

Offset

1,2

Comments

We have that Z_5*/(Z_5*)^n is the inverse limit of (Z/5^iZ)*/((Z/5^iZ)*)^n as i tends to infinity. Write n = 5^e * n' with n' not being divisible by 5, then the group is cyclic of order 5^e * gcd(4,n'). See A370050.

Links

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

Formula

Multiplicative with a(5^e) = 5^e, a(2) = 2, a(2^e) = 4 for e >= 2 and a(p^e) = 1 for primes p != 2, 5.

From Amiram Eldar, May 20 2024: (Start)

Dirichlet g.f.: (1 + 1/2^s + 1/2^(2*s-1)) * ((1 - 1/5^s)/(1 - 1/5^(s-1))) * zeta(s).

Sum_{k=1..n} a(k) ~ (8*n/(5*log(5))) * (log(n) + gamma - 1 + (3/4)*log(5/2)), where gamma is Euler's constant (A001620). (End)

Example

We have Z_5*/(Z_5*)^5 = Z_5* / ((1+25Z_5) U (7+25Z_5) U (18+25Z_5) U (24+25Z_5)) = (Z/25Z)*/((1+25Z) U (7+25Z) U (18+25Z) U (24+25Z)) = C_5, so a(5) = 5.
We have Z_5*/(Z_5*)^10 = Z_5* / ((1+25Z_5) U (24+25Z_5)) = (Z/25Z)*/((1+25Z) U (25+25Z)) = C_10, so a(10) = 10.

Mathematica

a[n_] := Module[{e2 = IntegerExponent[n, 2], e5 = IntegerExponent[n, 5]}, 2^Min[e2, 2] * 5^e5]; Array[a, 100] (* Amiram Eldar, May 20 2024 *)

Prog

(PARI) a(n, {p=5}) = my(e = valuation(n, p)); p^e*gcd(p-1, n/p^e)

Crossrefs

Row 3 of A370050. Cf. A001620, A297402, A370180, A370182.

Cf. A370566.

Sequence in context: A264017 A159971 A114158 * A248666 A162407 A368608

Adjacent sequences: A370178 A370179 A370180 * A370182 A370183 A370184

Keyword

nonn,easy,mult

Author

Jianing Song, Apr 30 2024

Status

approved