A319101 Number of solutions to x^7 == 1 (mod n).
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 7, 7
Offset: 1
Examples
Solutions to x^7 == 1 (mod 29): x == 1, 7, 16, 20, 23, 24, 25 (mod 29). Solutions to x^7 == 1 (mod 43): x == 1, 4, 11, 16, 21, 35, 41 (mod 43). Solutions to x^7 == 1 (mod 49): x == 1, 8, 15, 22, 29, 36, 43 (mod 49) (x == 1 (mod 7)).
Links
- Jianing Song, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Mathematica
f[p_, e_] := If[Mod[p, 7] == 1, 7, 1]; f[7, 1] = 1; f[7, e_] := 7; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Aug 10 2023 *)
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PARI
a(n)=my(Z=znstar(n)[2]); prod(i=1, #Z, gcd(7, Z[i]))
Formula
Multiplicative with a(7) = 1, a(7^e) = 7 if e >= 2; for other primes p, a(p^e) = 7 if p == 1 (mod 7), a(p^e) = 1 otherwise.
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(7, k_i).
Comments