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A319100
Number of solutions to x^6 == 1 (mod n).
15
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
1,3
COMMENTS
All terms are 3-smooth. a(n) is even for n > 2. Those n such that a(n) = 2 are in A066501.
LINKS
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.
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)*A060839(n).
For n > 2, a(n) = A060839(n)*2^A046072(n).
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.
a(n) = A000010(n)/A293483(n). - Jianing Song, Nov 10 2019
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
EXAMPLE
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).
PROG
(PARI) a(n)=my(Z=znstar(n)[2]); prod(i=1, #Z, gcd(6, Z[i]))
CROSSREFS
Number of solutions to x^k == 1 (mod n): A060594 (k=2), A060839 (k=3), A073103 (k=4), A319099 (k=5), this sequence (k=6), A319101 (k=7), A247257 (k=8).
Mobius transform gives A307381.
Sequence in context: A355192 A260983 A103222 * A304794 A175809 A061033
KEYWORD
nonn,easy,mult
AUTHOR
Jianing Song, Sep 10 2018
STATUS
approved