

A336013


Threecolumn 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
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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,n2) = 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^2y, x, print([x, y, (y^2y)/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



