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A273163
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P-defects p - N(p) of the congruence y^2 == x^3 + x^2 + 4*x + 4 (mod p) for primes p, where N(p) is the number of solutions given for p = prime(n) by A272207(n).
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5
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0, -2, -1, 2, 0, 2, -6, -4, 6, 6, -4, 2, 6, -10, -6, -6, 12, 2, 2, -12, 2, 8, 6, -6, 2, 6, 14, -6, 2, -6, 2, 0, 18, -4, -6, 20, -22, -10, 18, -6, -12, -10, -12, 26, 18, 8, -16, -10, -6, 14, -6, -24, 14, 0, -6, -18, 18, 20, 26, 6
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OFFSET
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1,2
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COMMENTS
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The modularity pattern series is the expansion of the 39th modular cusp form of weight 2 and level N = 20, given in Table I of the Martin reference, i.e., eta^2(2*z)*eta^2(10*z) = Sum_{n >= 1} c(n)*q^n in powers of q = exp(2*Pi*i*z), with Im(z) > 0, where eta is the Dedekind function. c(prime(n)) = a(n), also for the bad primes 2 and 5 (the discriminant of this elliptic curve is -2^4*5^2). See also A030205 for the expansion in powers of q^2 (after deleting the factor q^(-1/2)): A030205((prime(n)-1)/2) = a(n), n >= 2.
See the comment on the Martin-Ono reference in A272207 which implies that eta^2(2*z) * eta^2(10*z) provides the modularity sequence for this elliptic curve.
The elliptic curve y^2 = x^3 + x^2 - x has the same p-defects.
For the above defined q-series coefficients c one seems to have for primes not 2 and 5 c(prime(n)^k) = (sqrt(prime(n)))^k*S(k, a(n)/sqrt(prime(n))) with Chebyshev's S polynomials (A049310), for n = 2, and n >= 4 and k >= 2. This corresponds to alpha-multiplicativity with alpha(x) = x (weight 2) except for primes 2 and 5, obtainable formally from Sum_{n>=1, without factors 2 or 5} c(n)/n^s = Product_{n=2,n>=4} 1/(1 - a(n)/prime(n)^s + prime(n)/prime(n)^{2*s}) (absolute convergence of the product seems to hold for Re(s) > 1 needed for the prime 3 factor). For alpha-multiplicativity see the Apostol reference, pp. 138-139 (exercise 6). c(2*k) = 0, c(2*k+1) = A030205(k), k >= 0. Thus c(2^k) = c(2)^k = 0, and it seems that c(5^k) = A030205((5^k-1)/2) = c(5)^k = (-1)^k, for k >= 1. Then one has multiplicativity of {c(n)} with the given c(p^k) formula.
The Dirichlet sum could then be written as Sum_{n>=1} c(n)/n^s = 1/(1 - (-1)/5^s)* Product_{n=2,n>=4} 1/(1 - a(n)/prime(n)^s + prime(n)/prime(n)^{2*s}).
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LINKS
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FORMULA
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a(n) = prime(n) - A272207(n), n >= 1, where A272207(n) is the number of solutions to the congruence y^2 == x^3 + x^2 + 4*x + 4 (mod prime(n)).
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EXAMPLE
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{c(n)} multiplicativity test for n = 3^2*5^2 = 225: c(225) = A030205(112) = +1. c(3^2*5^2) = c(3^2)*c(5^2) = (3*S(2, a(2)/sqrt(3)))*(-1)^2 = ((-2)^2 - 3)*(+1) = +1.
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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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