A272207 Number of solutions to the congruence y^2 == x^3 + x^2 + 4*x + 4 (mod p) as p runs through the primes.
2, 5, 6, 5, 11, 11, 23, 23, 17, 23, 35, 35, 35, 53, 53, 59, 47, 59, 65, 83, 71, 71, 77, 95, 95, 95, 89, 113, 107, 119, 125, 131, 119, 143, 155, 131, 179, 173, 149, 179, 191, 191, 203, 167, 179, 191, 227, 233, 233, 215, 239, 263, 227, 251, 263, 281, 251, 251, 251, 275
Offset: 1
Examples
The first nonnegative complete residue system {0, 1, ..., prime(n)-1} is used. The solutions (x, y) of y^2 == x^3 + x^2 + 4*x + 4 (mod prime(n)) begin: n, prime(n), a(n)\ solutions (x, y) 1, 2, 2: (0, 0), (1, 0) 2, 3, 5: (0, 1), (0, 2), (1, 1), (1, 2) (2, 0) 3, 5, 6: (0, 2), (0, 3), (1, 0), (2, 2), (2, 3), (4, 0) 4, 7, 5: (0, 2), (0, 5), (4, 3), (4, 4), (6, 0) 5, 11, 11: (0, 2), (0, 9), (4, 1), (4, 10), (5, 3), (5, 8), (6, 4), (6, 7), (9, 5), (9, 6), (10, 0) ... The solutions (x, y) of y^2 == x^3 + x^2 - x (mod prime(n)) begin: n, prime(n), a(n)\ solutions (x, y) 1, 2, 2: (0, 0), (1, 1) 2, 3, 5: (0, 0), (1, 1), (1, 2), (2, 1) (2, 2) 3, 5, 6: (0, 0), (1, 1), (1, 4), (2, 0), (4, 1), (4, 4) 4, 7, 5: (0, 0), (1, 1), (1, 6), (6, 1), (6, 6) 5, 11, 11: (0, 0), (1, 1), (1, 10), (3, 0), (6, 2), (6, 9), (7, 0), (9, 3), (9, 8), (10, 1), (10, 10) ...
Links
- Seiichi Manyama, Table of n, a(n) for n = 1..10000
- Y. Martin, Multiplicative eta-quotients, Trans. Amer. Math. Soc. 348 (1996), no. 12, 4825-4856, see page 4852 Table I.
- Yves Martin and Ken Ono, Eta-Quotients and Elliptic Curves, Proc. Amer. Math. Soc. 125, No 11 (1997), 3169-3176.
Formula
a(n) gives the number of solutions of the congruence y^2 == x^3 + x^2 + 4*x + 4 (mod prime(n)), n >= 1.
a(n) gives also the number of solutions of the congruence y^2 == x^3 + x^2 - x (mod prime(n)), n >= 1.
Comments