A336013 -id:A336013 - cp's OEIS frontend

A332083 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.

-2, -1, -1, -1, -1, -2, 1, -1, 2, 2, -1, 1, -6, -2, -1, -3, -2, -2, -2, -2, -3, -1, -2, -6, 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, -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

Offset: 1

David Lovler, Aug 22 2020

All of the comments for A336013 also describe this table. Remember that id = -z/y and theta = -y/x.
Id and theta are both integers only for rows with x = +-1. - David Lovler, Feb 12 2022
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.
Proof that when id and theta are switched, f_1(a,b) and f_2(a,b) sum to a + b:
  Given id_2 = theta_1 and theta_2 = id_1,
  f_1(a,b) = (a*b - theta_1*(a+b) + id_1*theta_1)/(id_1 - theta_1)
  f_2(a,b) = (a*b - theta_2*(a+b) + id_2*theta_2)/(id_2 - theta_2)
  f_2(a,b) = (a*b - id_1*(a+b) + theta_1*id_1)/(theta_1 - id_1)
  f_2(a,b) = (-a*b + id_1*(a+b) - id_1*theta_1)/(id_1 - theta_1)
  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.
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].
Proof that if three rows pairwise sum to another row, then all three sum to [0, 1, 0]:
Lemma. If [x, y, z] is a row, then [0, 1, 0] - [x, y, z] is a row.
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.
Corollary. If [x, y, z] is not a row, then [0, 1, 0] - [x, y, z] is not a row
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.

			Table begins:
  [ x,  y,  z]
  -------------
  [-2, -1, -1];
  [-1, -1, -2];
  [ 1, -1,  2];
  [ 2, -1,  1];
  [-6, -2, -1];
  [-3, -2, -2];
  [-2, -2, -3];
  [-1, -2, -6];
  [ 1, -2,  6];
  [ 2, -2,  3];
  [ 3, -2,  2];
  [ 6, -2,  1];
  [-2,  2, -1];
  [-1,  2, -2];
  [ 1,  2,  2];
  [ 2,  2,  1];
  ...
Examples of two rows that sum to [0, 1, 0] and ids and thetas are switched:
  [x1, y1, z1] + [x2, y2, z2] = [2, 2, 1] + [-2, -1, -1] = [0, 1, 0],
    f_1(a,b) + f_2(a,b) = (2a*b + 2(a+b) + 1) + (-2a*b - (a+b) - 1) = a + b;
    id_1 = -z1/y1 = -1/2, theta_1 = -y1/x1 = -1,
    id_2 = -z2/y2 = -1 = theta_1, theta_2 = -y2/x2 = -1/2 = id_1;
    id_1/theta_2 + id_2/theta_1 = (-1/2)/(-1/2) + (-1)/(-1) = 2.
  [x1, y1, z1] + [x2, y2, z2] = [15, 6, 2] + [-15, -5, -2] = [0, 1, 0],
    f_1(a,b) + f_2(a,b) = (15a*b + 6(a+b) + 2) + (-15a*b - 5(a+b) - 2) = a + b;
    id_1 = -z1/y1 = -1/3, theta_1 = -y1/x1 = -2/5,
    id_2 = -z2/y2 = -2/5 = theta_1, theta_2 = -y2/x2 = -1/3 = id_1;
    id_1/theta_2 + id_2/theta_1 = (-1/3)/(-1/3) + (-2/5)/(-2/5) = 2.
Example of three rows that sum to a triple with y^2 - y - x*z = 0 and the rows pairwise sum to a row:
  [1, 7, 42] + [2, 8, 28] + [-3, -14, -70] = [0, 1, 0] and
  [1, 7, 42] + [2, 8, 28] = [3, 15, 70], another row;
  [1, 7, 42] + [-3, -14, -70] = [-2, -7, -28], another row;
  [2, 8, 28] + [-3, -14, -70] = [-1, -6, -42], another row.

David Lovler, Table of n, a(n) for n = 1..4368
David Lovler, The first 3912 triples for y up to 60.
David Lovler, Comments and examples

Cf. A336013.

{ my(y=1); fordiv (y^2+y, x, print([-(y^2+y)/x, -y, -x]) );
fordiv (y^2+y, x, print([x, -y, (y^2+y)/x]) );
for (y = 2, 6, fordiv (y^2+y, x, print([-(y^2+y)/x, -y, -x]) );
fordiv (y^2+y, x, print([x, -y, (y^2+y)/x]) );
fordiv (y^2-y, x, print([-(y^2-y)/x, y, -x]) );
fordiv (y^2-y, x, print([x, y, (y^2-y)/x]) ) )} \\ David Lovler, Apr 01 2021

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

A351581 Four-column table read by rows giving quadruples of integers [w,x,y,z] with w > 0, x > 1, y > 1 and z > 0 such that y^2 - y - xz = 0 and x^2 = wy, sorted by y then by x.

Original entry on oeis.org

2, 2, 2, 1, 3, 3, 3, 2, 12, 6, 3, 1, 1, 2, 4, 6, 4, 4, 4, 3, 9, 6, 4, 2, 36, 12, 4, 1, 5, 5, 5, 4, 20, 10, 5, 2, 80, 20, 5, 1, 6, 6, 6, 5, 150, 30, 6, 1, 7, 7, 7, 6, 28, 14, 7, 3, 63, 21, 7, 2, 252, 42, 7, 1, 2, 4, 8, 14, 8, 8, 8, 7, 98, 28, 8, 2, 392, 56, 8, 1

Offset: 1

David Lovler, Feb 13 2022

It is the same to sort by y then by w also to sort by y then by z descending.
When [w,x,y,z] is a row, f(a,b,c) = w*a*b*c + x*(a*b + a*c + b*c) + y*(a+b+c) + z is associative in the following sense. f((a,b,c),d,e) = f(a,f(b,c,d),e) = f(a,b,f(c,d,e)) for all a,b,c,d,e. f(a,b,c) is commutative because of its symmetry.
For each quadruple, the corresponding f(a,b,c) has a unique zero element (call it theta), meaning f(a,b,theta) = f(a,theta,b) = f(theta,a,b) = theta for all a,b. Theta = -y/x = - x/w. f(a,b,c) also has not one but two identity elements (id_1 and id_2), meaning f(a,id_1,id_1) = f(id_1,a,id_1) = f(id_1,id_1,a) = a for all a and f(a,id_2,id_2) = f(id_2,a,id_2) = f(id_2,id_2,a) = a for all a. Id = (-y +- sqrt(y))/x = theta +- sqrt(y)/x. Thus theta = (id_1 + id_2)/2.
The identity elements are integers when y is a square and x divides sqrt(y).

			Table begins:
  [ w,  x,  y,  z]
  -----------------
  [  2,  2, 2,  1];
  [  3,  3, 3,  2];
  [ 12,  6, 3,  1];
  [  1,  2, 4,  6];
  [  4,  4, 4,  3];
  [  9,  6, 4,  2];
  [ 36, 12, 4,  1];
  [  5,  5, 5,  4];
  [ 20, 10, 5,  2];
  [ 80, 20, 5,  1];
  [  6,  6, 6,  5];
  [150, 30, 6,  1];
  [  7,  7, 7,  6];
  [ 28, 14, 7,  3];
  [ 63, 21, 7,  2];
  [252, 42, 7,  1];
  [  2,  4, 8, 14];
  [  8,  8, 8,  7];
  [ 98, 28, 8,  2];
  [392, 56, 8,  1];
  [  1,  3, 9, 24];
  [  4,  6, 9, 12];
  [  9,  9, 9,  8];
  [ 16, 12, 9,  6];
  [ 36, 18, 9,  4];
  [ 64, 24, 9,  3];
  [144, 36, 9,  2];
  [576, 72, 9,  1];
  ...
For row [1, 2, 4, 6], f(a,b,c) = a*b*c + 2*(a*b + a*c + b*c) + 4*(a+b+c) + 6. Theta = -2; id_1 = -1, id_2 = -3. The associative function f(a,b) = a*b + 2*(a+b) + 2 has theta = -2 and id = -1; f(f(a,b),c) = f(a,b,c). Another associative function g(a,b) = -a*b - 2*(a+b) - 6 with theta = -2 and id = -3 likewise gives g(g(a,b),c) = f(a,b,c).

David Lovler, Table of n, a(n) for n = 1..2720
David Lovler, The first 680 quadruples for y up to 100.

Cf. A336013, A351986.

{ my(y); for (y = 2, 9, fordiv (y^2-y, x, if(type(w = x^2/y) == "t_INT", print([w, x, y, (y^2-y)/x]) )) ) }

Looking at A336013, if [X,Y,Z] is a row and f(a,b) = X*a*b + Y*(a+b) + Z is the corresponding associative function with id = -Z/Y and theta = -Y/X, then the composition f(f(a,b),c) = w*a*b*c + x*(a*b + a*c + b*c) + y*(a+b+c) + z = f(a,b,c) gives the quadruple [w,x,y,z]. f(a,b,c) has the same theta as f(a,b); the two identity elements for f(a,b,c) are id and 2*theta - id.
If theta and the identity elements are computed from a quadruple, f(a,b,c) can be written as (a*b*c - theta*(a*b + a*c + b*c) + theta^2*(a+b+c) - theta^3)/(id-theta)^2 + theta. The square in the denominator ensures that f(a,b,c) is the same for either id.
Two parameters are sufficient to describe a row. For n*s > 1, rows are [w,x,y,z] = [n, n*s, n*s^2, (n^2*s^4-n*s^2)/(n*s)] = [n, n*s, n*s^2, n*s^3 - s]. In terms of n and s, theta = -s and id = s*(-1 +- 1/sqrt(n)). Rows with s=1 stand out as having w=x=y; theta = -1 and id = -1 +- 1/sqrt(w).

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A332083 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.

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A351581 Four-column table read by rows giving quadruples of integers [w,x,y,z] with w > 0, x > 1, y > 1 and z > 0 such that y^2 - y - xz = 0 and x^2 = wy, sorted by y then by x.

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Programs

PARI

Formula

cp's OEIS Frontend

A332083 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.

Views

Author

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Comments

Examples

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Crossrefs

Programs

PARI

Formula

A351581 Four-column table read by rows giving quadruples of integers [w,x,y,z] with w > 0, x > 1, y > 1 and z > 0 such that y^2 - y - x*z = 0 and x^2 = w*y, sorted by y then by x.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A351581 Four-column table read by rows giving quadruples of integers [w,x,y,z] with w > 0, x > 1, y > 1 and z > 0 such that y^2 - y - xz = 0 and x^2 = wy, sorted by y then by x.