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%I #19 Nov 26 2016 03:03:12
%S 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,
%T 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,
%U 0,0,-28,26,0
%N 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).
%C 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.
%C 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).
%C 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.
%C 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.
%C 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, ...
%D 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).
%H Seiichi Manyama, <a href="/A272198/b272198.txt">Table of n, a(n) for n = 1..10000</a>
%H Y. Martin, <a href="http://dx.doi.org/10.1090/S0002-9947-96-01743-6">Multiplicative eta-quotients</a>, Trans. Amer. Math. Soc. 348 (1996), no. 12, 4825-4856, see page 4852 Table I.
%H Yves Martin and Ken Ono, <a href="http://www.ams.org/journals/proc/1997-125-11/S0002-9939-97-03928-2/">Eta-Quotients and Elliptic Curves</a>, Proc. Amer. Math. Soc. 125, No 11 (1997), 3169-3176.
%F 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)).
%F a(n) = 0 for prime(n) == 0, 2 (mod 3) (see A045309).
%F 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.
%F a(n) = +2*A001479(m+1) if prime(n) == A002476(m) (m is unique) is a prime of A272200 (type I).
%F a(n) = -2*A001479(m+1) if prime(n) == A002476(m) is from A272201 (type II).
%F See a comment above for the bisection of the primes 1 (mod 3) into type I and II.
%e a(1) = 2 - A272197(1) = 0, and 2 == 2(mod 3).
%e 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.
%e 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.
%Y Cf. A000040, A002476, A001479, A001480, A045309, A272197, A272200, A272201.
%K sign,easy
%O 1,4
%A _Wolfdieter Lang_, May 02 2016
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