cp's OEIS Frontend

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

For n=1 one has the Modd 1 residue class [0], the integers. The group of order 1 is the cyclic group Z_1 with the unit element 0==1 (Modd 1). [Changed by Wolfdieter Lang, Apr 04 2012]

For the non-cyclic (acyclic) values see A206552.

For these numbers n, and only for these (only the n values < 100 are shown above), there exist primitive roots Modd n. See the nonzero values of A206550 for the smallest positive ones.

For n=1 the primitive root is 0 == 1 (Modd 1), see above.

For n>=1 the multiplicative group Modd n 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.

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.

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.

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).