A370566 Size of the group Q_5*/(Q_5*)^n, where Q_5 is the field of 5-adic numbers.

1, 4, 3, 16, 25, 12, 7, 32, 9, 100, 11, 48, 13, 28, 75, 64, 17, 36, 19, 400, 21, 44, 23, 96, 625, 52, 27, 112, 29, 300, 31, 128, 33, 68, 175, 144, 37, 76, 39, 800, 41, 84, 43, 176, 225, 92, 47, 192, 49, 2500, 51, 208, 53, 108, 275, 224, 57, 116, 59, 1200, 61, 124, 63, 256

Offset: 1

Author

Jianing Song, Apr 30 2024

Keywords

• nonn
• easy
• mult

Comments

We have Q_5* = 5^Z X Z_5*, so Q_5*/(Q_5*)^k = (5^Z/5^(kZ)) X (Z_p*/(Z_5*)^k). Note that 5^Z/5^(kZ) is a cyclic group of order k. For the group structure of (Z_5*/(Z_5*)^k), see A370050.

Links

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

Crossrefs

Row 3 of A370067. Cf. A001620, A370050, A370564, A370565, A370567.

Cf. A370181.

Programs

Mathematica

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

PARI

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

Formula

Write n = 5^e * n' with k' not being divisible by 5, then a(n) = n * 5^e * gcd(4,n').

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

a(n) = n * A370181(n).

From Amiram Eldar, May 20 2024: (Start)

Dirichlet g.f.: ((1 + 1/2^(s-1) + 1/2^(2*s-3)) * (1 - 1/5^(s-1))/(1 - 1/5^(s-2))) * zeta(s-1).

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