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A278720
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The p-defect p - N(p) of the congruence y^2 == x^3 + 4*x (mod p) for primes p, where N(p) is the number of solutions given by A276730.
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4
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0, 0, -2, 0, 0, 6, 2, 0, 0, -10, 0, -2, 10, 0, 0, 14, 0, -10, 0, 0, -6, 0, 0, 10, 18, -2, 0, 0, 6, -14, 0, 0, -22, 0, 14, 0, 22, 0, 0, -26, 0, -18, 0, -14, -2, 0, 0, 0, 0, 30, 26, 0, -30, 0, 2, 0, -26, 0, -18, 10, 0, -34, 0, 0, 26, 22, 0, 18, 0, -10, 34, 0, 0, 14, 0, 0, -34, 38, 2, -6, 0, 30, 0, 34, 0, 0, -14, 42, 38, 0, 0, 0, 0, 0, 0, 0, -10, -22, 0, -42
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OFFSET
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1,3
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COMMENTS
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This sequence gives also the p-defects for the congruences y^2 == x^3 - x (mod p), y^2 == x^3 - 11*x - 14 (mod p) and y^2 == x^3 - 11*x + 14 (mod p). See the Cremona link, Table 1, N = 32. - Wolfdieter Lang, Dec 22 2016
This elliptic curve y^2 = x^3 + 4*x appears as strong Weil curve for the weight 2 newform (eta(4*tau)*eta(8*tau))^2 of level N=32, with Dedekind's eta function. See the Martin-Ono link, Theorem 2, p. 3173, the row with Conductor 32. See also A002171 for the expansion of this newform in powers of q^4 (but with different offset). The also Nr. 49 of the Martin Table 1.
From this L-series of this elliptic curve one has:
a(n) = 0 if prime(n) == 2 or 3 (mod 4). (see the conjecture by Robert Israel, Sep 28 2016 in A276730).
If prime(n) == 1 (mod 4) = A002144(m) (for a unique m = m(n)) then prime(n) = A(m)^2 + B(m)^2 with the odd A(m) = A002972(m) and the even B(m) = 2*A002973(m). It turns out that 4*A002144(m) - a(m^2) = (2*B(m))^2 for m=m(n), and the sign s(m) of a(m) is + if A(m) + B(m) == 1 (mod 4) and - if A(m) + B(m) == 3 (mod 4). For the primes == 1 (mod 4) leading to sign + or - see A279392 or A279393, respectively. One has thus s(m) = (-1)^((A(m)-1)/2 + B(m)/2). See the Martin-Ono formula for a_{32}(p) in Theorem 3, p. 3175. This leads to the a(n) formula given below.
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LINKS
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FORMULA
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a(n) = 0 if prime(n) == 2 or 3 (mod 4) (this is conjecture II from above).
a(n) = s(m)*2*A(m) if prime(n) = A002144(m), with A(m) = A002972(m) and the sign s(m) = (-1)^((A(m)-1)/2 + B(m)/2).
a(n) = - Sum_{k=1..p-3} ((k*(k+1)*(k+2))/p) where (x/y) is the Kronecker symbol. - Michel Marcus, Nov 06 2020
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EXAMPLE
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a(1) = 0 because prime(1) = 2 == 2 (mod 4).
a(2) = 0 because prime(2) = 3 == 3 (mod 4).
a(3) = -2 because prime(3) = 5 = A002144(1) = A002972(1)^2 + (2*A002973(1))^2 = 1^2 + 2^2. Hence 2*A(1) = 2*A002972(1) = 2, and the sign s(1) = - because A(1) + B(1) = 1 + 2*1 = 3 == 3 (mod 4).
a(6) = +6 because prime(6) = 13 = A002144(2) = A(2)^2 + B(2)^2 = 3^2 + (2*1)^2. Hence 2*A(2) = 6 and the sign is + because A(2) + B(2) = 3 + 2 = 5 == 1 (mod 4).
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PROG
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(PARI) a(n) = my(p=prime(n)); -sum(k=1, p-3, kronecker(k*(k+1)*(k+2), p)); \\ Michel Marcus, Nov 06 2020
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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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