cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A272198 The p-defect p - N(p) of the congruence y^2 == x^3 + 1 (mod p) for primes p, where N(p) is the number of solutions given by A272197(n).

Original entry on oeis.org

0, 0, 0, -4, 0, 2, 0, 8, 0, 0, -4, -10, 0, 8, 0, 0, 0, 14, -16, 0, -10, -4, 0, 0, 14, 0, 20, 0, 2, 0, 20, 0, 0, -16, 0, -4, 14, 8, 0, 0, 0, 26, 0, 2, 0, -28, -16, -28, 0, -22, 0, 0, 14, 0, 0, 0, 0, -28, 26, 0
Offset: 1

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Author

Wolfdieter Lang, May 02 2016

Keywords

Comments

This sequence for an elliptic curve (of the Bachet-Mordell type) is discussed in the Silverman reference. In Exercise 45.5, in the table on p. 405, the p-defects are called a_p, and are shown for primes 2 to 113.
The modularity pattern series is the expansion of the 51st modular cusp form of weight 2 and level N=36, given in the table I of the Martin reference, i.e., eta^4(6*z) in powers of q = exp(2*Pi*i*z), with Im(z) > 0. Here eta is the Dedekind function. See A000727 for the expansion in powers of q^6 (after deleting a factor q^(1/6)). Note that also for the possibly bad prime 2 and the bad prime 3 this expansion gives the correct numbers 0 (the discriminant of this elliptic curve is -3^3).
See also the comment on the Martin-Ono reference in A272197 which implies that eta^4(6*z) provides the modularity sequence for this elliptic curve.
For primes p == 0 and 2 (mod 3) (A045309) a(p) = 0. The proof runs along the same line as the one given in the Silverstein reference on pp. 400 - 402 for 17 replaced by 1. From the expansion of the known modularity function eta^4(6*z) follows that only the coefficients for powers q^n with n == 1 (mod 6) are nonzero, and therefore all a(p) for primes p == 0 and 2 (mod 3) have to vanish.
If prime(n) == 1 (mod 3) = A002476(m) (for a unique m = m(n)) then prime(n) = A(m)^2 + 3*B(m)^2 with A(m) = A001479(m+1) and B(m) = A001480(m+1), m >= 1. In this case (4*prime(n) - a(n)^2)/12, seems to be a square, q(m)^2. In fact is seems that (the positive) q(m) = B(m). This is true at least for the first 80 primes 1 (mod 3), i.e. for such primes <= 997. (In the Silverman reference, in hint c) for Exercise 4.5, on p. 405, a more complicated way is suggested: 4*p is decomposed there non-uniquely instead of p uniquely.) If this conjecture is true then a(n) = 2*(+/-sqrt(prime(n) - 3*B(m)^2)) = +- 2*A(m) for prime(n) = A002476(m). This leads to a bisection of the primes 1 (mod 3) into two types: type I if the + sign applies, and type II for the - sign. Primes of type I are given in A272200: 13, 19, 43, 61, 97, ... and those of type II in A272201: 7, 31, 37, 67, 73, ...

Examples

			a(1) = 2 - A272197(1) = 0, and 2 == 2(mod 3).
a(4) = 7 - A272197(4) = 7 - 11 = -4, and 7 = A002476(1) = 2^2 + 3*1^2, 2 = A001479(1+1), 7 = A272201(1), hence a(4) = -2*2 = -4.
a(6) = 13 - A272197(6) = 13 - 11 = 2, and 13 = A002476(2) = 1^2 + 3*2^2; 1 = A001479(2+1), 13 = A272200(1), hence a(6) = +2*1 = +2.
		

References

  • J. H. Silverman, A Friendly Introduction to Number Theory, 3rd ed., Pearson Education, Inc, 2006, Exercise 45.5, p. 405, Exercise 47.2, p. 415, and pp. 400 - 402 (4th ed., Pearson 2014, Exercise 5, p. 371, Exercise 2, p. 385, and pp. 366 - 368).

Crossrefs

Formula

a(n) = prime(n) - N(prime(n)), n = 1, where N(prime(n)) = A272197(n), the number of solutions of the congruence y^2 == x^3 + 1 (mod prime(n)).
a(n) = 0 for prime(n) == 0, 2 (mod 3) (see A045309).
The above given conjecture for primes 1 (mod 3) is true because Mordell proved the Ramanujan conjecture on the expansion coefficients of eta^4(6*z), and with the present a(n) the result of Ramanujan follows. See the references and a comment on A000727.
a(n) = +2*A001479(m+1) if prime(n) == A002476(m) (m is unique) is a prime of A272200 (type I).
a(n) = -2*A001479(m+1) if prime(n) == A002476(m) is from A272201 (type II).
See a comment above for the bisection of the primes 1 (mod 3) into type I and II.

A272201 Bisection of primes congruent to 1 modulo 3 (A002476), depending on the corresponding A001479 entry being congruent to 1 modulo 3 or not. Here the second case.

Original entry on oeis.org

7, 31, 37, 67, 73, 79, 139, 151, 199, 211, 223, 229, 271, 307, 313, 337, 367, 397, 421, 439, 457, 541, 547, 571, 577, 613, 643, 709, 739, 751, 823, 829, 853, 877, 907, 919, 997
Offset: 1

Views

Author

Wolfdieter Lang, Apr 28 2016

Keywords

Comments

The other primes congruent to 1 modulo 3 are given in A272200, where also more details are given.
Each prime == 1 (mod 3) has a unique representation A002476(m) = A(m)^2 + 3*B(m)^2 with positive A(m) = A001479(m+1) and B(m) = A001480(m+1), m >= 1 (see also A001479). The present sequence gives these primes corresponding to A(m+1) not congruent 1 modulo 3. The ones corresponding to A(m+1) == 1 (mod 3) (the complement) are given in A272200.

Crossrefs

Cf. A000727, A001479, A002476, A001480, A272198, A272200 (complement relative to A002476).

Programs

  • Maple
    filter:= proc(n) local S,x,y;
        if not isprime(n) then return false fi;
        S:= remove(hastype,[isolve(x^2+3*y^2=n)],negative);
        subs(S[1],x) mod 3 <> 1
    end proc:
    select(filter, [seq(i,i=7..1000,6)]); # Robert Israel, Apr 29 2019
  • Mathematica
    filterQ[n_] := Module[{S, x, y}, If[!PrimeQ[n], Return[False]]; S = Solve[x > 0 && y > 0 && x^2 + 3 y^2 == n, Integers]; Mod[x /. S[[1]], 3] != 1];
    Select[Range[7, 1000, 6], filterQ] (* Jean-François Alcover, Apr 21 2020, after Robert Israel *)

Formula

This sequence collects the 1 (mod 3) primes p(m) = A002476(m) = A(m)^2 + 3*B(m)^2 with positive A(m) not == 1 (mod 3), for m >= 1. A(m) = A001479(m+1).

A272202 Number of solutions of the congruence y^2 == x^3 - 1 (mod p) as p runs through the primes.

Original entry on oeis.org

2, 3, 5, 3, 11, 11, 17, 27, 23, 29, 27, 47, 41, 51, 47, 53, 59, 47, 51, 71, 83, 75, 83, 89, 83, 101, 123, 107, 107, 113, 147, 131, 137, 123, 149, 147, 143, 171, 167, 173, 179, 155, 191, 191, 197, 171, 195, 195, 227, 251, 233, 239, 227, 251, 257, 263, 269, 243, 251, 281
Offset: 1

Views

Author

Wolfdieter Lang, May 05 2016

Keywords

Comments

In the Martin and Ono reference, in Theorem 2, this elliptic curve appears in the last column, starting with Conductor 144, as a strong Weil curve for the weight 2 newform eta^{12}(12*z) / (eta^4(6*z) * eta^4(24*z)), symbolically 12^{12} 6^{-4} 24^{-4}, with Im(z) > 0, and the Dedekind eta function. See A187076 which gives the q-expansion (q = exp(2*Pi*i*z)) of exp(-Pi*i*z/3)* eta(2*z)^{12} / (eta^4(z)*eta^4(4*z)). For the q-expansion of 12^{12} 6^{-4} 24^{-4} one has a leading zero and 5 interspersed 0's: 0,1,0,0,0,0,0,4,0,0,0,0,0,2,0,0,0,0,0,-8,...
The discriminant of this elliptic curve is -3^3 = -27.
For the elliptic curve y^2 == x^3 + 1 (mod prime(n)) see A000727, A272197, A272198, A272200 and A272201.

Examples

			The first nonnegative complete residue system {0, 1, ..., prime(n)-1} is used. The solutions (x, y) of y^2 == x^3 - 1 (mod prime(n)) begin:
n, prime(n), a(n)\ solutions (x, y)
1,    2,      2:   (0, 1), (1, 0)
2,    3,      3:   (1, 0), (2, 1), (2, 2)
3,    5,      5:   (0, 2), (0, 3), (1, 0),
                   (3, 1), (3, 4)
4,    7,      3:   (1, 0), (2, 0), (4, 0)
5,    11,    11:   (1, 0), (3, 2), (3, 9),
                   (5, 5), (5, 6), (7, 1),
                   (7, 10), (8, 4), (8, 7),
                   (10, 3), (10, 8)
...
		

Crossrefs

Formula

a(n) gives the number of solutions of the congruence y^2 == x^3 - 1 (mod prime(n)), n >= 1.
Showing 1-3 of 3 results.