%I
%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 pdefect 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 BachetMordell type) is discussed in the Silverman reference. In Exercise 45.5, in the table on p. 405, the pdefects 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 MartinOno 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 nonuniquely 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/S0002994796017436">Multiplicative etaquotients</a>, Trans. Amer. Math. Soc. 348 (1996), no. 12, 48254856, see page 4852 Table I.
%H Yves Martin and Ken Ono, <a href="http://www.ams.org/journals/proc/199712511/S0002993997039282/">EtaQuotients and Elliptic Curves</a>, Proc. Amer. Math. Soc. 125, No 11 (1997), 31693176.
%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
