|
%I #19 May 20 2024 02:31:20
%S 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,
%T 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,
%U 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
%N Size of the group Z_5*/(Z_5*)^n, where Z_5 is the ring of 5-adic integers.
%C 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.
%H Jianing Song, <a href="/A370181/b370181.txt">Table of n, a(n) for n = 1..10000</a>
%F 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.
%F From _Amiram Eldar_, May 20 2024: (Start)
%F Dirichlet g.f.: (1 + 1/2^s + 1/2^(2*s-1)) * ((1 - 1/5^s)/(1 - 1/5^(s-1))) * zeta(s).
%F 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)
%e 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.
%e 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.
%t 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 *)
%o (PARI) a(n,{p=5}) = my(e = valuation(n, p)); p^e*gcd(p-1, n/p^e)
%Y Row 3 of A370050. Cf. A001620, A297402, A370180, A370182.
%Y Cf. A370566.
%K nonn,easy,mult
%O 1,2
%A _Jianing Song_, Apr 30 2024
|
