A319100 Number of solutions to x^6 == 1 (mod n).
1, 1, 2, 2, 2, 2, 6, 4, 6, 2, 2, 4, 6, 6, 4, 4, 2, 6, 6, 4, 12, 2, 2, 8, 2, 6, 6, 12, 2, 4, 6, 4, 4, 2, 12, 12, 6, 6, 12, 8, 2, 12, 6, 4, 12, 2, 2, 8, 6, 2, 4, 12, 2, 6, 4, 24, 12, 2, 2, 8, 6, 6, 36, 4, 12, 4, 6, 4, 4, 12, 2, 24, 6, 6, 4, 12, 12, 12, 6, 8, 6, 2
Offset: 1
Examples
Solutions to x^6 == 1 (mod 13): x == 1, 3, 4, 9, 10, 12 (mod 13). Solutions to x^6 == 1 (mod 27): x == 1, 8, 10, 17, 19, 26 (mod 27) (x == 1, 8 (mod 9)). Solutions to x^6 == 1 (mod 37): x == 1, 10, 11, 26, 27, 36 (mod 37).
Links
- Jianing Song, Table of n, a(n) for n = 1..10000
- Steven Finch, Greg Martin and Pascal Sebah, Roots of unity and nullity modulo n, Proc. Amer. Math. Soc., Vol. 138, No. 8 (2010), pp. 2729-2743.
Crossrefs
Programs
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PARI
a(n)=my(Z=znstar(n)[2]); prod(i=1, #Z, gcd(6, Z[i]))
Formula
Multiplicative with a(2) = 1, a(4) = 2, a(2^e) = 4 for e >= 3; a(3) = 2, a(3^e) = 6 if e >= 2; for other primes p, a(p^e) = 6 if p == 1 (mod 6), a(p^e) = 2 if p == 5 (mod 6).
If the multiplicative group of integers modulo n is isomorphic to C_{k_1} x C_{k_2} x ... x C_{k_m}, where k_i divides k_j for i < j; then a(n) = Product_{i=1..m} gcd(6, k_i).
a(n) = A060594(n) iff n is not divisible by 9 and no prime factor of n is congruent to 1 mod 6, that is, n in A088232.
Sum_{k=1..n} a(k) ~ c * n * log(n)^3, where c = (1/Pi^4) * Product_{p prime == 1 (mod 6)} (1 - (12*p-4)/(p+1)^3) = 0.0075925601... (Finch et al., 2010). - Amiram Eldar, Mar 26 2021
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