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%I #34 Dec 08 2024 04:00:44
%S 1,-1,-2,0,-3,-1,1,2,1,2,-1,-2,-1,-3,1,0,6,1,0,3,2,1,-1,-3,1,-2,-2,-2,
%T -1,0,-1,0,-1,0,-1,0,0,6,-2,0,6,-2,0,6,-2,1,4,1,6,-1,2,2,2,3,-2,-1,-5,
%U 4,3,1,-2,-4,-5,-3,-1,1,0,-6,-1,0,-3,-2,0,-6,-1,0,-3,-2,1,-2,-7,0,2,-2,-1,0,-5,-2,-5,3,4,-1,2,3,-2,2,4,2,-2,1,6,-1,4,2,4
%N Triangle T(n, k) read by rows. Let m be a nonzero rational number then T(n, m mod (n+1)) is the n-th coefficient in the Hasse-Weil L-series (q^(n+1) in the q-expansion) associated to the elliptic equation -4*x^3 + ((m+1)^2 + 8)*x^2 - 2*(m+3)*x + 1 - y^2 = 0.
%C Unfortunately, if m is a fraction m = b/c, this triangle can only be used for those coefficients where c and (n+1) are coprime. This is not only because the modulo operation is undefined otherwise, but also because rows of the triangle where (n+1) divides c contain these coefficients with the wrong sign.
%C The parametrization model for elliptic equations defined by -4*x^3 + ((m+1)^2 + 8)*x^2 - 2*(m+3)*x + 1 - y^2 is also used in A377441. From its relation to Somos-4 sequences, it is known that there is at least one generator point of infinite order if m is an integer > 0 or < -1. If we assume the Birch and Swinnerton-Dyer conjecture to be true, then we expect the associated L-function L(E, s) to be zero at s = 1 for such m.
%C The relation of m to the J-invariant is given by J(m) = (m^12 + 12*m^11 + 114*m^10 + 628*m^9 + 2823*m^8 + 8184*m^7 + 19036*m^6 + 24552*m^5 + 25407*m^4 + 16956*m^3 + 9234*m^2 + 2916*m + 729)/(m^5 + 4*m^4 + 23*m^3 + 9*m^2) for rational m.
%C The row sums of the triangle show some connection to the Dedekind psi function (A001615), but will deviate for at least many nonsquarefree n+1.
%C A short table which shows the Cremona label which corresponds to the L-series obtained for some rational m:
%C .
%C m | label
%C -------------
%C -5 655a1
%C -4 166a1
%C -3 153a1
%C -2 58a1
%C -1 11a3
%C -1/2 26b1
%C -1/3 141a1
%C 1 37a1
%C 2 158b1
%C 3 423g1
%C 4 458a1
%C 5 1745b1
%C .
%F T(n, n) = A006571(n), case m =-1. Also the expansion of (eta(q) * eta(q^11))^2 in powers of q.
%F T(n, 1) = A007653(n), case m = 1.
%F T(2*n, n) = A251913(2*n+1), case m = -1/2. See first comment.
%F Let p be an odd prime with good reduction, then T(p-1, k) is odd iff -4*x^3 + ((k+1)^2 + 8)*x^2 - 2*(k+3)*x + 1 == 0 (mod p) has no solution.
%e The triangle T(n, k) begins:
%e q^(n+1) 0, 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 sum A001615
%e --------------------------------------------------------------------
%e [q^1] 1 1 1
%e [q^2] -1,-2 -3 3
%e [q^3] 0,-3,-1 -4 4
%e [q^4] 1, 2, 1, 2 6 6
%e [q^5] -1,-2,-1,-3, 1 6 6
%e [q^6] 0, 6, 1, 0, 3, 2 12 12
%e [q^7] 1,-1,-3, 1,-2,-2,-2 -8 8
%e [q^8] -1, 0,-1, 0,-1, 0,-1, 0 -4 12 <- not equal
%e [q^9] 0, 6,-2, 0, 6,-2, 0, 6,-2 12 12
%e [q^10] 1, 4, 1, 6,-1, 2, 2, 2, 3,-2 18 18
%e [q^11] -1,-5, 4, 3, 1,-2,-4,-5,-3,-1, 1 12 12
%e [q^12] 0,-6,-1, 0,-3,-2, 0,-6,-1, 0,-3,-2 -24 24
%e [q^13] 1,-2,-7, 0, 2,-2,-1, 0,-5,-2,-5, 3, 4 -14 14
%e [q^14] -1, 2, 3,-2, 2, 4, 2,-2, 1, 6,-1, 4, 2, 4 24 24
%e [q^15] 0, 6, 1, 0,-3, 1, 0, 3, 3, 0, 3, 2, 0, 9,-1 24 24
%e [q^16] 1,-4, 1,-4, 1,-4, 1,-4, 1,-4, 1,-4, 1,-4, 1,-4 24 24
%o (PARI)
%o T(n, k) = ellak(ellinit(ellfromeqn(-4*x^3 + ((k+n+2)^2 + 8)*x^2 - 2*(k+n+4)*x + 1 - y^2)),n+1);
%Y Cf. A001615, A251913, A377441.
%Y Cf. A006571 (main diagonal), A007653 (column 1).
%K sign,tabl
%O 0,3
%A _Thomas Scheuerle_, Nov 17 2024