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A336013 Three-column table read by rows giving triples of integers with x > 0, y > 1 and z > 0 such that y^2 - y - x*z = 0, sorted by y then by x.

Original entry on oeis.org

1, 2, 2, 2, 2, 1, 1, 3, 6, 2, 3, 3, 3, 3, 2, 6, 3, 1, 1, 4, 12, 2, 4, 6, 3, 4, 4, 4, 4, 3, 6, 4, 2, 12, 4, 1, 1, 5, 20, 2, 5, 10, 4, 5, 5, 5, 5, 4, 10, 5, 2, 20, 5, 1, 1, 6, 30, 2, 6, 15, 3, 6, 10, 5, 6, 6, 6, 6, 5, 10, 6, 3, 15, 6, 2, 30, 6, 1
Offset: 1

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Author

David Lovler, Jul 07 2020

Keywords

Comments

When [x, y, z] is a row, f(a,b) = x*a*b + y*(a+b) + z is associative.
For each triple, the corresponding f(a,b) has an identity element (id), meaning f(a,id) = f(id,a) = a for all a. Id = -z/y. f(a,b) also has a zero element (call it theta), meaning f(a,theta) = f(theta,a) = theta for all a. Theta = -y/x.
I and θ are both integers only for rows with x = 1. - David Lovler, Feb 12 2022
f(a,b), defined by each row, also has a distributive rule when the generalized zero is taken into account. This means that if we define a "partition" of b by b = b1 + b2 - theta, then f(a,b) = f(a,b1 + b2 - theta) = f(a,b1) + f(a,b2) - theta for all a and b, and all "partitions" of b. Notice that when theta = 0, we have the usual distributive rule.
Another way to write f(a,b) for a row is to first compute id and theta from x, y and z. Then f(a,b) = (a*b - theta*(a+b) + id*theta)/(id - theta); or equivalently f(a,b) = (a*b - theta*(a+b) + theta^2)/(id - theta) + theta. Notice that id cannot equal theta because of id - theta in the denominator. Also notice that when id = 1 and theta = 0, f(a,b) = a*b, but multiplication is not represented in the table since the corresponding row would be [1, 0, 0], which is not allowed.
If (i) two rows are [x1, y1, z1] and [x2, y2, z2],
(ii) id_1 = -z1/y1, id_2 = -z2/y2, theta_1 = -y1/x1, theta_2 = -y2/x2, and
(iii) id_1/theta_2 + id_2/theta_1 = 2,
then [x1+x2, y1+y2, z1+z2] is a row. Consequently, if
(i) f1(a,b) = x1*a*b + y1*(a+b) + z1 is associative and
f2(a,b) = x2*a*b + y2*(a+b) + z2 is associative,
(ii) id_1 = -z1/y1, id_2 = -z2/y2, theta_1 = -y1/x1, theta_2 = -y2/x2, and
(iii) id_1/theta_2 + id_2/theta_1 = 2,
then f1(a,b) + f2(a,b) = (x1+x2)*a*b + (y1+y2)*(a+b) + z1+z2 is associative. Proof: Given [x1, y1, z1] and [x2, y2, z2] are rows and id_1/theta_2 + id_2/theta_1 = 2, the following algebraic manipulations show that [x1+x2, y1+y2, z1+z2] is a row.
(-z1/y1)/(-y2/x2) + (-z2/y2)/(-y1/x1) = 2
(x2*z1)/(y1*y2) + (x1*z2)/(y1*y2) = 2 [Multiply by y1*y2 and move to the right.]
0 = 2*y1*y2 - x1*z2 - x2*z1
[Add to the right side y1^2 - y1 - x1*z1 and y2^2 - y2 - x2*z2 which are both 0.]
0 = 2*y1*y2 - x1*z2 - x2*z1 + y1^2 - y1 - x1*z1 + y2^2 - y2 - x2*z2 [Rearrange.]
0 = (y1^2 + 2*y1*y2 + y2^2) - y1 - y2 - x1*z1 - x1*z2 - x2*z1 - x2*z2
0 = (y1+y2)^2 - (y1+y2) - (x1+x2)*(z1+z2). QED.
The idea of summing rows to get another row can be extended. If (i) three rows are [x1, y1, z1], [x2, y2, z2], and [x3, y3, z3], (ii) id and theta are defined as above, and (iii) (id_1/theta_2 + id_2/theta_1 - 2)/y3 + (id_1/theta_3 + id_3/theta_1 - 2)/y2 + (id_2/theta_3 + id_3/theta_2 - 2)/y1 = 0, then [x1+x2+x3, y1+y2+y3, z1+z2+z3] is a row.
Generalizing, when summing n rows to another row, the criterion involves the sum of binomial(n,2) versions of id_i/theta_j + id_j/theta_i - 2 as i and j go from 1 to n and i < j. Furthermore, each of these expressions is divided by the product of the y values from rows other than i and j. There are binomial(n,n-2) = binomial(n,2) such products. Formally this is:
If [x1,y1,z1] , ..., [xn,yn,zn] are rows and Sum_{1<=i
All of the above comments are still true when x, y, z, id and theta are complex numbers with y != 1 and x, y, z != 0.
If [x, y, z] is not a row, compute K = y/(y^2 - x*z). Then, K*[x, y, z] is a row if it is an integer triple. Note that if [x, y, z] were a row, K = 1. Furthermore, if [n*x, n*y, n*z] is not a row, compute K' = n*y/((n*y)^2 - (n*x)*(n*z)) = y/(n*(y^2 - x*z)) = K/n. Then K'*[n*x, n*y, n*z] = K*[x, y, z] as before. When K*[x, y, z] is not a triple for not having integer values, we still have (K*y)^2 - K*y - (K*x)*(K*z) = 0. - David Lovler, Jan 24 2022

Examples

			Table begins:
  [ x, y,  z]
-------------
  [ 1, 2,  2];
  [ 2, 2,  1];
  [ 1, 3,  6];
  [ 2, 3,  3];
  [ 3, 3,  2];
  [ 6, 3,  1];
  [ 1, 4, 12];
  [ 2, 4,  6];
  [ 3, 4,  4];
  [ 4, 4,  3];
  [ 6, 4,  2];
  [12, 4,  1];
  [ 1, 5, 20];
  [ 2, 5, 10];
  [ 4, 5,  5];
  [ 5, 5,  4];
  [10, 5,  2];
  [20, 5,  1];
  ...
Example of the distributive rule:
  [x, y, z] = [1, 2, 2]
  f(a,b) = a*b + 2*(a+b) + 2
  theta = -y/x = -2
  f(5,7) = 35 + 2*(5+7) + 2 = 61 which equals
  f(5,3 + 2 -(-2)) = f(5,3) + f(5,2) - (-2) = (15 + 16 + 2) + (10 + 14 + 2) + 2 = 61.
Examples of rows that sum to another row:
  [1, 7, 42] + [2, 8, 28] = [3, 15, 70] because
  id_1/theta_2 + id_2/theta_1 = (-42/7)/(-8/2) + (-28/8)/(-7/1) = 2.
  [42, 7, 1] + [28, 8, 2] = [70, 15, 3] because
  id_1/theta_2 + id_2/theta_1 = (-1/7)/(-8/28) + (-2/8)/(-7/42) = 2.
  [2, 8, 28] + [3, 18, 102] = [5, 26, 130] because
  id_1/theta_2 + id_2/theta_1 = (-28/8)/(-18/3) + (-102/18)/(-8/2) = 2.
  [28, 8, 2] + [102, 18, 3] = [130, 26, 5] because
  id_1/theta_2 + id_2/theta_1 = (-2/8)/(-18/102) + (-3/18)/(-8/28) = 2.
Examples of three rows that sum to a row:
  [1, 2, 2] + [1, 2, 2] + [1, 5, 20] = [3, 9, 24] because
  (id_1/theta_2 + id_2/theta_1 - 2)/y3 + (id_1/theta_3 + id_3/theta_1 - 2)/y2 + (id_2/theta_3 + id_3/theta_2 - 2)/y1 = ((-2/2)/(-2/1) + (-2/2)/(-2/1) - 2)/5 + ((-2/2)/(-5/1) + (-20/5)/(-2/1) - 2)/2 + ((-2/2)/(-5/1) + (-20/5)/(-2/1) - 2)/2 = 0. In this example no two of the rows sum to another row.
  [1, 7, 42] + [2, 8, 28] + [3, 10, 30] = [6, 25, 100] because
  (id_1/theta_2 + id_2/theta_1 - 2)/y3 + (id_1/theta_3 + id_3/theta_1 - 2)/y2 + (id_2/theta_3 + id_3/theta_2 - 2)/y1 = ((-42/7)/(-8/2) + (-28/8)/(-7/1) - 2)/10 + ((-42/7)/(-10/3) + (-30/10)/(-7/1) - 2)/8 + ((-28/8)/(-10/3) + (-30/10)/(-8/2) - 2)/7 = 0. In this example [1, 7, 42] and [2, 8, 28] sum to [3, 15, 70], another row.
For n=4,
  if (id_1/theta_2 + id_2/theta_1 - 2)/(y3*y4) + (id_1/theta_3 + id_3/theta_1 - 2)/(y2*y4) + (id_1/theta_4 + id_4/theta_1 - 2)/(y2*y3) + (id_2/theta_3 + id_3/theta_2 - 2)/(y1*y4) + (id_2/theta_4 + id_4/theta_2 - 2)/(y1*y3) + (id_3/theta_4 + id_4/theta_3 - 2)/(y1*y2) = 0,
  then [x1+x2+x3+x4 , y1+y2+y3+y4 , z1+z2+z3+z4] is a row.
[x, y, z] = [3, 2, 1] is not a row, but (y/(y^2 - x*z))*[x, y, z] = (2/(2^2 - 3*1))*[3, 2, 1] = [6, 4, 2] which is a row. - _David Lovler_, Jan 22 2022

Links

Crossrefs

The number of rows for each y beginning with y=2 is A092517.
Cf. A332083.

Programs

  • PARI
    for (y = 2, 8, fordiv (y^2-y, x, print([x, y, (y^2-y)/x]) ) ) \\ David Lovler, Mar 12 2021

Formula

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

Extensions

Table format in example edited by David Lovler, Feb 14 2022

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