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A111725
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Number of residues modulo n of the maximum order.
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10
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1, 1, 1, 1, 2, 1, 2, 3, 2, 2, 4, 3, 4, 2, 4, 4, 8, 2, 6, 4, 6, 4, 10, 7, 8, 4, 6, 6, 12, 4, 8, 8, 12, 8, 8, 6, 12, 6, 8, 8, 16, 6, 12, 12, 8, 10, 22, 8, 12, 8, 16, 8, 24, 6, 16, 14, 18, 12, 28, 8, 16, 8, 24, 16, 24, 12, 20, 16, 30, 8, 24, 14, 24, 12, 16, 18, 24, 8, 24, 24, 18, 16, 40, 14, 32, 12
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OFFSET
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1,5
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COMMENTS
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The maximum order modulo n is given by A002322(n).
a(n) is the number of primitive lambda-roots of n. - Michel Marcus, Mar 17 2016
A primitive lambda-root is an element of maximal order modulo n. - Joerg Arndt, Mar 19 2016
a(n) is odd if and only if n is a factor of 24, i.e., n is in A018253. - Jianing Song, Apr 27 2019
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LINKS
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FORMULA
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Decompose (Z/nZ)* as a product of cyclic groups C_{k_1} x C_{k_2} x ... x C_{k_m}, where k_i divides k_j for i < j, then a(n) = Sum_{d divides psi(n)} (mu(psi(n)/d)*Product{i=1..m} gcd(d, k_i)). This is an immediate corollary from the fact that the number of elements in (Z/nZ)* such that x^d == 1 (mod n) is Product{i=1..m} gcd(d, k_i). Here (Z/nZ)* is the multiplicative group of integers modulo n, psi(n) = A002322(n) and mu(n) = A008683(n). - Jianing Song, Apr 27 2019
Decompose (Z/nZ)* as a product of cyclic groups C_{k_1} x C_{k_2} x ... x C_{k_m}, where k_i divides k_j for i < j, then a(n) = phi(n) * Product_{p prime dividing phi(n)} (1 - 1/p^r(p)), where r(p) is the number of j such that v(k_j,p) = v(k_m,p), v(s,p) is the p-adic valuation of s.
Proof: let G = (Z/nZ)*, G_p be the Sylow p-subgroup of G, then G = Product_{p prime dividing phi(n)} G_p: every element x can be uniquely written as Product_{p prime dividing phi(n)} x_p for x_p in G_p. Let ord(x) be the order of x. Since ord(x_p, x_p') = 1 for distinct p and p', we have ord(x) = Product_{p prime dividing phi(n)} ord(x_p), hence x is of maximal order if and only if each x_p is of maximal order in G_p.
Each G_p is isomorphic to C_{p^(e_1)} x C_{p^(e_2)} x ... x C_{p^(e_m)} for e_1 <= e_2 <= ... <= e_m, e_m > 0. Write x_p = (x_{p,1}, x_{p,2}, ..., x_{p,m}). Suppose that e_m = e_{m-1} = ... = e_{m-r+1} > e_{m-r}, then x_p is of maximal order in G_p if and only of x_{p,j} is of order p^(e_m) for some m-r+1 <= j <= m, so the number of such x_p is p^(e_1) * p^(e_2) * ... * p^(e_{m-r}) * (p^(r*e_m) - p^(r*((e_m)-1))) = |G_p| * (1 - 1/p^r).
An example: n = 15903, then (Z/nZ)* = C_6 x C_18 x C_90. We can see that r(2) = 3, r(3) = 2 and r(5) = 1, so a(15903) = phi(15903) * (1 - 1/2^3) * (1 - 1/3^2) * (1 - 1/5^1) = 6048.
It should be clear that a(n) = phi(phi(n)) if and only if r(p) = 1 for every prime p dividing phi(n), or v(k_{m-1},p) < v(k_m,p) for every prime p dividing phi(n). Otherwise, a(n) > phi(phi(n)). (End)
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MAPLE
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LiDelta := proc(q, n)
local a, p, e, lam, v ;
a := 0 ;
lam := numtheory[lambda](n) ;
for p in numtheory[factorset](n) do
e := padic[ordp](n, p) ;
if p =2 and e= 3 and q =2 and padic[ordp](lam, q) = 1 then
elif isprime(q) then
v := padic[ordp](lam, q) ;
if modp( numtheory[lambda](p^e), q^v) = 0 then
a := a+1 ;
end if;
end if:
end do:
a ;
end proc:
local a, q ;
a := 1;
for q in numtheory[factorset](numtheory[lambda](n)) do
a := a*(1-1/q^LiDelta(q, n)) ;
end do:
a*numtheory[phi](n) ;
end proc:
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MATHEMATICA
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f[list_]:=Count[list, Max[list]]; Map[f, Table[Table[MultiplicativeOrder[k, n], {k, Select[Range[n], GCD[#, n]==1&]}], {n, 1, 100}]] (* Geoffrey Critzer, Jan 26 2013 *)
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PROG
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(PARI) { a(n) = my(r, c, r1); r=1; c=0; for(k=0, n-1, if(gcd(k, n)!=1, next); r1=znorder(Mod(k, n)); if(r1==r, c++); if(r1>r, r=r1; c=1) ); c; }
(PARI) { A111725(n) = if(n<3, return(1)); my(k, p); k=znstar(n)[2]; p=factor(k[1])[, 1]; eulerphi(n) * prod(i=1, #p, (1-1/p[i]^vecsum(apply(x->valuation(k[1]\x, p[i])==0, k))) ); } \\ Max Alekseyev, Oct 23 2021
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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