

A273163


Pdefects 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).


5



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

1,2


COMMENTS

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 MartinOno 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 pdefects.
For the above defined qseries 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 alphamultiplicativity 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 alphamultiplicativity see the Apostol reference, pp. 138139 (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^k1)/2) = c(5)^k = (1)^k, for k >= 1. Then one has multiplicativity of {c(n)} with the given c(p^k) formula.
For this multiplicity see the Michael Somos Oct 31 2005 comment on A030205, where c(n) is called b(n).  Wolfdieter Lang, May 24 2016
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}).


LINKS

Seiichi Manyama, Table of n, a(n) for n = 1..10000


FORMULA

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)).


EXAMPLE

{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.


CROSSREFS

Cf. A000040, A030205, A049310, A272207.
Sequence in context: A137423 A127471 A240312 * A276695 A071485 A127969
Adjacent sequences: A273160 A273161 A273162 * A273164 A273165 A273166


KEYWORD

sign,easy


AUTHOR

Wolfdieter Lang, May 20 2016


STATUS

approved



