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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. 1
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

When [x, y, z] is a row of the table, 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.

f(a,b), defined by each row of the table, 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 of the table 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 of the table 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 of the table. 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 of the table 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 of the table.

(-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 of the table to get another row of the table can be extended. If (i) three rows of the table 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 of the table.

Generalizing, when summing n rows to another row of the table, 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 of the table and Sum_{1<=i<j<=n} ((id_i/theta_j + id_j/theta_i - 2) / (Product_{k=1..n and k!=i,j} y_k)) = 0, then [Sum_{m=1..n} x_m , Sum_{m=1..n} y_m , Sum_{m=1..n} z_m] is a row of the table.

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.

LINKS

Table of n, a(n) for n=1..78.

FORMULA

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

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

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

EXAMPLE

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 of the table 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 of the table that sum to a row of the table:

[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 of the table.

[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 of the table.

PROG

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

CROSSREFS

The number of rows of the table for each y beginning with y=2 is A092517.

Cf. A332083.

Sequence in context: A037200 A263992 A180174 * A172363 A181877 A236472

Adjacent sequences:  A336010 A336011 A336012 * A336014 A336015 A336016

KEYWORD

nonn,tabf

AUTHOR

David Lovler, Jul 07 2020

STATUS

approved

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Last modified June 19 07:26 EDT 2021. Contains 345126 sequences. (Running on oeis4.)