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For multiplication Modd n (not to be confused with mod n) see a comment on A203571.

The 0 for n=1 is a primitive root Modd 1, the other zeros indicate that there is no primitive root for this n.

Iff a(n)>0, for n>=2, then the Galois group Gal(Q(2*cos(Pi/n))/Q), which is the multiplicative group of odd reduced residue classes Modd n (hence the notation Modd) is cyclic. For n=1 this group is also cyclic. See A206551 (cyclic moduli n) and A206552 (acyclic, i.e. non-cyclic, moduli n). [Changed by Wolfdieter Lang, Apr 04 2012]

For Modd n (not to be confused with mod n) see a comment on A203571.

Precisely these numbers n (only the ones <=165 are shown above) have no primitive root Modd n. See the zero entries of A206550, except A206550(1) = 0 which stands for a primitive root 0.

The multiplicative Modd n group is the Galois group Gal(Q(rho(n))/Q), with the algebraic number rho(n) := 2*cos(Pi/n) with minimal polynomial C(n,x), whose coefficients are given in A187360.

A055034(n) is the degree delta(n) of the minimal polynomial of the algebraic number rho(n):=2*cos(pi/n), n>=1, whose coefficients are shown in A187360. It is also the order of multiplicative abelian group Modd n (for multiplication Modd n see a comment on A203571). This is the Galois group Gal(Q(rho(n))/Q). If the number of abelian groups of order delta(n) is 1 then this group is necessarily cyclic.

Because A000688 is 1 exactly for the squarefree numbers A005117, the set of a(n) values of the present sequence is a (proper) subset of A206551. Hence it is immediately clear that the multiplicative group Modd a(n) is cyclic, but there are other cyclic Modd n groups, e.g., for n = 8, 10, 15, 16, 17, 19, 26, 27, 32, 34, 35, 37, 38, 39, 41,...

If n belongs to A206551 (cyclic multiplicative group Modd n) then there exist precisely a(n) primitive roots Modd n. For these n values the number of entries in row n of the table A216319 with value delta(n) (the row length) is a(n). Note that a(n) is also defined for the complementary n values from A206552 (non-cyclic multiplicative group Modd n) for which no primitive root Modd n exists.

See also A216322 for the number of primitive roots Modd n.

The number of digits of a(n) is 1, 1, 5, 17, 54, 167, 506, 1525, ... .

Integer 2*a(n) implies that 5^delta(3^n) == -1 (mod 3^n), n>=1, with the degree delta(3^n) = phi(2*3^n)/2 = 3^(n-1) of the minimal polynomial C(3^n,x) of the algebraic number 2*cos(pi/3^n). For delta and the coefficient array of C see A055034 and A187360, respectively. That 2*a(n) is indeed an even integer can be shown by analyzing the terms of the binomial expansion of (6-1)^(3^(n-1)) + 1.

This congruence implies that floor(5^(3^(n-1))/3^n) = 2*a(n) - 1, i.e., it is odd. Hence 5^delta(3^n) == +1 (Modd 3^n), n>=2. For Modd n (not to be confused with mod n) see a comment on A203571. One can show that 5 is the smallest positive primitive root Modd 3^n for n>=2 (for n=1 one has 1^1 == +1 (Modd 3)). See A206550. The proof uses the fact that the order of 5 using multiplication Modd 3^n has to be a divisor of delta(3^n)=3^(n-1), i.e., a power of 3. This is because the multiplicative group Modd 3^n has order delta(3^n) and the subgroup formed by the cycle has the order of 5 considered Modd 3^n. Then apply Lagrange's theorem. That for n>=2 no number 5^(3^(n-1-j)), with j=1, 2..., n-1, is congruent +1 (Modd 3^n) follows from the above established congruence and an analysis of the relevant expansion for a given smaller power.

The above statements show that for n>=1 the multiplicative group Modd 3^n is cyclic (for n=1 the cycle is [1], and for n>=2 the cycle is generated by 5). For the cyclic moduli see A206551.

The number of digits of a(n) is 1, 4, 22, 117, 593, 2978, 14905, ..., for n >= 1.

Integer a(n) implies that 3^delta(5^n) == -1 (mod 5^n), n>=1, with the degree delta(5^n) = phi(2*5^n)/2 = 2*5^(n-1) of the minimal polynomial C(5^n,x) of the algebraic number 2*cos(Pi/5^n). For delta and the coefficient array of C see A055034 and A187360, respectively. That 2*a(n) is indeed an even integer can be shown by analyzing the terms of the binomial expansion of (10-1)^(5^(n-1)) + 1.

This congruence implies that floor(3^(2*5^(n-1))/5^n) = 2*a(n) - 1, i.e., it is odd. Hence 3^delta(5^n) == +1 (Modd 5^n), n>=1. For Modd n (not to be confused with mod n) see a comment on A203571. One can show that 3 is the smallest positive primitive root Modd 5^n for n>=1. See A206550. The proof uses the fact that the order of 3 using multiplication Modd 5^n has to be a divisor of delta(5^n)=2*5^(n-1), i.e., either 5^e or 2*5^e with certain e>=0. This is because the multiplicative group Modd 5^n has order delta(5^n) and the order of the subgroup formed by the cycle generated by 3 coincides with the order of 3 considered Modd 5^n. Then Lagrange's theorem is applied. That for n>=1 no power of 3 with exponent 2*5^(n-1-j) with j=1,2,..., n-1 is congruent +1 (Modd 5^n) follows by considering the two cases +1 (mod 5^n) and -1 (mod 5^n) separately. The first case is excluded from the above established congruence by an indirect proof. The case -1 (Modd 5^n) can be excluded by an analysis of the relevant expansion for a given smaller power. The other cases 3^k with k = 5^(n-1-j), where j = 0, 1, ..., n-1, are neither -1 (mod 5^n) nor +1 (mod 5^n) because 3^(5^(n-1)) (mod 5^n) is congruent 3 (mod 5) (see A048899), hence neither +1 nor -1 (mod 5^n), respectively. The lower exponents are then excluded in both cases iteratively by an indirect proof taking fifth powers.

The above statements show that for n>=1 the multiplicative group Modd 5^n is cyclic, and for each n the cycle of length 2*5^(n-1) can be generated starting with 3. For the cyclic moduli see A206551.

This sequence coincides with A216321 for all n values from A206551 (cyclic multiplicative Modd n group) and the entry is 0 otherwise (if no primitive root exists, that is, n is from the complementary sequence A206552).

For Modd n (not to be confused with mod n) see a comment on A203571.

The row lengths sequence for this array is 1 for row no. 1 and (p(n)-1)/2 with p(n):=A000040(n) (the primes).

For the definition of the index of a reduced number a mod n (but here we use Modd n) relative to a primitive root mod n, see, e.g., the Apostol reference, p. 213, and the tables on pp. 216-7. This mod n array is found under A054503 if the smallest primitive root mod n is taken as base. Because of its properties the index ind_b(a) is also called log_b(a), with the base b.

Here for Modd n, n>=2, primitive roots exist only for the values n with A206550(n)>0. There the smallest positive primitive roots, called here B(n) are also found. The allowed n values are shown in A206551. The indices Modd p(n), p(n):=A000040(n) (the primes) are called Ind_B(p(n))(a), with the odd numbers a smaller than p(n): 2*m-1=1,3,...,p(n)-2, for m=1,2,..., (p(n)-1)/2.

For odd p(n) the index Ind_B(p(n))(2*m-1) is defined as the unique value k from {0,1,...,(p(n)-3)/2}, such that B(p(n))^k = 2*m-1, with the base B(p(n)) the smallest positive primitive root Modd p(n).

The degree delta(m) of the minimal polynomial of rho(m) := 2*cos(Pi/m), called C(m,x) with coefficient array A187360, is given by A055034(m).

If delta(m) = phi(2*m)/2, m>=2, delta(1) = 1, with phi = A000010, is prime then the (Abelian) Galois group G(Q(rho(m))/Q) is cyclic. Because this Galois group of C(m,x) has order delta(m) this follows from a corollary to Lagrange's theorem, or also from Cauchy's theorem on groups.

Because the mentioned Galois group is isomorphic to the multiplicative group Modd m of order delta(m) (see a comment on A203571) all m = a(n) values appear in A206551.

This sequence is also a subsequence of A210845 because p is squarefree (see A005117).

This sequences differs from A206551 because no Factor Pair Latin Square of order 45 exists.

It is known that every prime power (A000961) as well as twice every prime number (A100484) appears in this sequence.