A332083 - OEIS

%I #63 Mar 05 2022 01:20:30

%S -2,-1,-1,-1,-1,-2,1,-1,2,2,-1,1,-6,-2,-1,-3,-2,-2,-2,-2,-3,-1,-2,-6,

%T 1,-2,6,2,-2,3,3,-2,2,6,-2,1,-2,2,-1,-1,2,-2,1,2,2,2,2,1,-12,-3,-1,-6,

%U -3,-2,-4,-3,-3,-3,-3,-4,-2,-3,-6,-1,-3,-12,1,-3,12,2,-3,6,3,-3,4,4,-3,3,6,-3,2,12,-3,1

%N Three-column table read by rows giving triples of integers with x, y, z != 0 and y != 1 such that y^2 - y - x*z = 0, sorted by the absolute value of y with the negatives first, then by x in ascending order.

%C All of the comments for A336013 also describe this table. Remember that id = -z/y and theta = -y/x.

%C Id and theta are both integers only for rows with x = +-1. - _David Lovler_, Feb 12 2022

%C There exist pairs of rows [x1, y1, z1] and [x2, y2, z2] for which id_2 = theta_1 and theta_2 = id_1 making id_1/theta_2 + id_2/theta_1 = 1 + 1 = 2. They provide the simplest instances of summing two rows and getting a triple for which y^2 - y - x*z = 0. In these cases the sum of the two rows is always [0, 1, 0] which is not in the table but corresponds to 0*a*b + 1*(a+b) + 0 = a + b.

%C Proof that when id and theta are switched, f_1(a,b) and f_2(a,b) sum to a + b:

%C   Given id_2 = theta_1 and theta_2 = id_1,

%C   f_1(a,b) = (a*b - theta_1*(a+b) + id_1*theta_1)/(id_1 - theta_1)

%C   f_2(a,b) = (a*b - theta_2*(a+b) + id_2*theta_2)/(id_2 - theta_2)

%C   f_2(a,b) = (a*b - id_1*(a+b) + theta_1*id_1)/(theta_1 - id_1)

%C   f_2(a,b) = (-a*b + id_1*(a+b) - id_1*theta_1)/(id_1 - theta_1)

%C   f_1(a,b) + f_2(a,b) = (a*b - a*b + (id_1 - theta_1)*(a+b) + id_1*theta_1 -  id_1*theta_1)/(id_1 - theta_1) = a + b. QED.

%C With the introduction of negative x, y and z we have the possibility of summing three rows to a triple for which y^2 - y - x*z = 0, and each pair of the three rows sums to a row. This was not possible in A336013. However when this happens, the result is not a row. This is because in these cases, the sum of three rows is always [0, 1, 0].

%C Proof that if three rows pairwise sum to another row, then all three sum to [0, 1, 0]:

%C Lemma. If [x, y, z] is a row, then [0, 1, 0] - [x, y, z] is a row.

%C Proof of lemma. [0, 1, 0] - [x, y, z] = [-x, 1-y, -z]. Rename this triple [X, Y, Z] and show that Y^2 - Y - X*Z = 0. Y^2 - Y - X*Z = (1-y)^2 - (1-y) - (-x)*(-z) = 1 - 2y + y^2 - 1 + y - x*z = y^2 - y - x*z = 0.

%C Corollary. If [x, y, z] is not a row, then [0, 1, 0] - [x, y, z] is not a row

%C Given rows [x1, y1, z1] and [x2, y2, z2] that sum to another row. By the lemma, [0, 1, 0] - ([x1, y1, z1] + [x2, y2, z2]) = a row, call it [x3, y3, z3]. Rearranging the equation first so that [x1, y1, z1] alone is on the right side we see that [x2, y2, z2] + [x3, y3, z3] is a row, and second so that [x2, y2, z2] alone is on the right side we see that [x1, y1, z1] + [x3, y3, z3] is a row. Finally, moving all the triples to the right side we get [0, 1, 0] = [x1, y1, z1] + [x2, y2, z2] + [x3, y3, z3]. QED.

%H David Lovler, <a href="/A332083/b332083.txt">Table of n, a(n) for n = 1..4368</a>

%H David Lovler, <a href="/A332083/a332083_1.txt">The first 3912 triples for y up to 60</a>.

%H David Lovler, <a href="/A332083/a332083_1.pdf">Comments and examples</a>

%F x = (y^2 - y)/z.

%F y = (1 +- sqrt(1 + 4*x*z))/2.

%F z = (y^2 - y)/x.

%e Table begins:

%e   [ x,  y,  z]

%e   -------------

%e   [-2, -1, -1];

%e   [-1, -1, -2];

%e   [ 1, -1,  2];

%e   [ 2, -1,  1];

%e   [-6, -2, -1];

%e   [-3, -2, -2];

%e   [-2, -2, -3];

%e   [-1, -2, -6];

%e   [ 1, -2,  6];

%e   [ 2, -2,  3];

%e   [ 3, -2,  2];

%e   [ 6, -2,  1];

%e   [-2,  2, -1];

%e   [-1,  2, -2];

%e   [ 1,  2,  2];

%e   [ 2,  2,  1];

%e   ...

%e Examples of two rows that sum to [0, 1, 0] and ids and thetas are switched:

%e   [x1, y1, z1] + [x2, y2, z2] = [2, 2, 1] + [-2, -1, -1] = [0, 1, 0],

%e     f_1(a,b) + f_2(a,b) = (2a*b + 2(a+b) + 1) + (-2a*b - (a+b) - 1) = a + b;

%e     id_1 = -z1/y1 = -1/2, theta_1 = -y1/x1 = -1,

%e     id_2 = -z2/y2 = -1 = theta_1, theta_2 = -y2/x2 = -1/2 = id_1;

%e     id_1/theta_2 + id_2/theta_1 = (-1/2)/(-1/2) + (-1)/(-1) = 2.

%e   [x1, y1, z1] + [x2, y2, z2] = [15, 6, 2] + [-15, -5, -2] = [0, 1, 0],

%e     f_1(a,b) + f_2(a,b) = (15a*b + 6(a+b) + 2) + (-15a*b - 5(a+b) - 2) = a + b;

%e     id_1 = -z1/y1 = -1/3, theta_1 = -y1/x1 = -2/5,

%e     id_2 = -z2/y2 = -2/5 = theta_1, theta_2 = -y2/x2 = -1/3 = id_1;

%e     id_1/theta_2 + id_2/theta_1 = (-1/3)/(-1/3) + (-2/5)/(-2/5) = 2.

%e Example of three rows that sum to a triple with y^2 - y - x*z = 0 and the rows pairwise sum to a row:

%e   [1, 7, 42] + [2, 8, 28] + [-3, -14, -70] = [0, 1, 0] and

%e   [1, 7, 42] + [2, 8, 28] = [3, 15, 70], another row;

%e   [1, 7, 42] + [-3, -14, -70] = [-2, -7, -28], another row;

%e   [2, 8, 28] + [-3, -14, -70] = [-1, -6, -42], another row.

%o (PARI) { my(y=1); fordiv (y^2+y, x, print([-(y^2+y)/x, -y, -x]) );

%o fordiv (y^2+y, x, print([x, -y, (y^2+y)/x]) );

%o for (y = 2, 6, fordiv (y^2+y, x, print([-(y^2+y)/x, -y, -x]) );

%o fordiv (y^2+y, x, print([x, -y, (y^2+y)/x]) );

%o fordiv (y^2-y, x, print([-(y^2-y)/x, y, -x]) );

%o fordiv (y^2-y, x, print([x, y, (y^2-y)/x]) ) )} \\ _David Lovler_, Apr 01 2021

%Y Cf. A336013.

%K sign,tabf

%O 1,1

%A _David Lovler_, Aug 22 2020