%I
%S 0,0,0,4,0,2,0,8,0,0,4,10,0,8,0,0,0,14,16,0,10,4,0,0,14,0,20,0,2,
%T 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,
%U 28,26,0,32,0,16,0,22,0,32,34,0,14,0,0,4,38,8,0,0,34,0,38,0,22,0,2,28,0,0,10,0,20,0,0,44,0,32,0,0,0,8,46,40,0,0,0,16,46
%N Pdefects 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 A272202(n).
%C The analysis of this elliptic curve runs along the same lines as in A000727, A272197 and A272198, and it is inspired by the Silverman reference where the curve y^2 = x^3 + 1 modulo primes is treated.
%C The series showing the modularity pattern is the expansion of the 67th modular cusp form of weight 2 and level N=144, given in the table I of the Martin reference, i.e., eta^{12}(12*z)/( eta^4(6*z)*eta^4(24*z)), symbolically 12^{12} 6^(4) 24^{4}, in powers of q = exp(2*Pi*i*z), with Im(z) > 0. Here eta is the Dedekind function. See A187076 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 A272202 which implies that 12^{12} 6^(4) 24^{4} provides the modularity sequence for this elliptic curve.
%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). 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 A272204: 7,13,31,61,67, ... and those of type II in A272205: 19,37,43,73,103, ...
%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="/A272203/b272203.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://dx.doi.org/10.1090/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)) = A272202(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 expected to be true by analogy to the case A272198 where only the signs differ.
%F a(n) = +2*A001479(m+1) if prime(n) == A002476(m) (m is unique) is a prime of A272204 (type I).
%F a(n) = 2*A001479(m+1) if prime(n) == A002476(m) is from A272205 (type II).
%F See a comment above for this bisection of the primes 1 (mod 3) into type I and II.
%e a(1) = 2  A272202(1) = 0, and 2 == 2 (mod 3).
%e a(4) = 7  A272202(4) = 7  3 = +4, and 7 = A002476(1) = 2^2 + 3*1^2, 2 = A001479(1+1), 7 = A272204(1), hence a(4) = +2*2 = +4.
%e a(8) = 19  A272202(8) = 19  27 = 8, and 19 = A002476(3) = 4^2 + 3*1^2; 4=A001479(3+1), 19 = A272205(1), hence a(8) = 2*(4) = 8.
%Y Cf. A000040, A187076, A272202, A272204, A272205.
%K sign,easy
%O 1,4
%A _Wolfdieter Lang_, May 05 2016
