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A272198
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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).
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6
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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
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OFFSET
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1,4
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COMMENTS
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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, ...
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REFERENCES
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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).
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LINKS
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FORMULA
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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.
See a comment above for the bisection of the primes 1 (mod 3) into type I and II.
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EXAMPLE
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a(1) = 2 - A272197(1) = 0, and 2 == 2(mod 3).
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CROSSREFS
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KEYWORD
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sign,easy
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AUTHOR
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STATUS
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approved
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