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
KEYWORD
nonn,easy,mult
AUTHOR
Jianing Song, Apr 30 2024
STATUS
approved