This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A336013 #80 Feb 15 2022 16:41:08
%S A336013 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,
%T A336013 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,
%U A336013 5,6,6,6,6,5,10,6,3,15,6,2,30,6,1
%N 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.
%C A336013 When [x, y, z] is a row, f(a,b) = x*a*b + y*(a+b) + z is associative.
%C A336013 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.
%C A336013 I and θ are both integers only for rows with x = 1. - _David Lovler_, Feb 12 2022
%C A336013 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.
%C A336013 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.
%C A336013 If (i) two rows are [x1, y1, z1] and [x2, y2, z2],
%C A336013 (ii) id_1 = -z1/y1, id_2 = -z2/y2, theta_1 = -y1/x1, theta_2 = -y2/x2, and
%C A336013 (iii) id_1/theta_2 + id_2/theta_1 = 2,
%C A336013 then [x1+x2, y1+y2, z1+z2] is a row. Consequently, if
%C A336013 (i) f1(a,b) = x1*a*b + y1*(a+b) + z1 is associative and
%C A336013 f2(a,b) = x2*a*b + y2*(a+b) + z2 is associative,
%C A336013 (ii) id_1 = -z1/y1, id_2 = -z2/y2, theta_1 = -y1/x1, theta_2 = -y2/x2, and
%C A336013 (iii) id_1/theta_2 + id_2/theta_1 = 2,
%C A336013 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.
%C A336013 (-z1/y1)/(-y2/x2) + (-z2/y2)/(-y1/x1) = 2
%C A336013 (x2*z1)/(y1*y2) + (x1*z2)/(y1*y2) = 2 [Multiply by y1*y2 and move to the right.]
%C A336013 0 = 2*y1*y2 - x1*z2 - x2*z1
%C A336013 [Add to the right side y1^2 - y1 - x1*z1 and y2^2 - y2 - x2*z2 which are both 0.]
%C A336013 0 = 2*y1*y2 - x1*z2 - x2*z1 + y1^2 - y1 - x1*z1 + y2^2 - y2 - x2*z2 [Rearrange.]
%C A336013 0 = (y1^2 + 2*y1*y2 + y2^2) - y1 - y2 - x1*z1 - x1*z2 - x2*z1 - x2*z2
%C A336013 0 = (y1+y2)^2 - (y1+y2) - (x1+x2)*(z1+z2). QED.
%C A336013 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.
%C A336013 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:
%C A336013 If [x1,y1,z1] , ..., [xn,yn,zn] are rows 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.
%C A336013 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.
%C A336013 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
%H A336013 David Lovler, <a href="/A336013/b336013.txt">Table of n, a(n) for n = 1..2898</a>
%H A336013 David Lovler, <a href="/A336013/a336013.txt">The first 8492 triples for y up to 300</a>
%H A336013 David Lovler, <a href="/A336013/a336013.pdf">Comments and examples</a>
%F A336013 x = (y^2 - y)/z.
%F A336013 y = (1 + sqrt(1 + 4*x*z))/2.
%F A336013 z = (y^2 - y)/x.
%e A336013 Table begins:
%e A336013 [ x, y, z]
%e A336013 -------------
%e A336013 [ 1, 2, 2];
%e A336013 [ 2, 2, 1];
%e A336013 [ 1, 3, 6];
%e A336013 [ 2, 3, 3];
%e A336013 [ 3, 3, 2];
%e A336013 [ 6, 3, 1];
%e A336013 [ 1, 4, 12];
%e A336013 [ 2, 4, 6];
%e A336013 [ 3, 4, 4];
%e A336013 [ 4, 4, 3];
%e A336013 [ 6, 4, 2];
%e A336013 [12, 4, 1];
%e A336013 [ 1, 5, 20];
%e A336013 [ 2, 5, 10];
%e A336013 [ 4, 5, 5];
%e A336013 [ 5, 5, 4];
%e A336013 [10, 5, 2];
%e A336013 [20, 5, 1];
%e A336013 ...
%e A336013 Example of the distributive rule:
%e A336013 [x, y, z] = [1, 2, 2]
%e A336013 f(a,b) = a*b + 2*(a+b) + 2
%e A336013 theta = -y/x = -2
%e A336013 f(5,7) = 35 + 2*(5+7) + 2 = 61 which equals
%e A336013 f(5,3 + 2 -(-2)) = f(5,3) + f(5,2) - (-2) = (15 + 16 + 2) + (10 + 14 + 2) + 2 = 61.
%e A336013 Examples of rows that sum to another row:
%e A336013 [1, 7, 42] + [2, 8, 28] = [3, 15, 70] because
%e A336013 id_1/theta_2 + id_2/theta_1 = (-42/7)/(-8/2) + (-28/8)/(-7/1) = 2.
%e A336013 [42, 7, 1] + [28, 8, 2] = [70, 15, 3] because
%e A336013 id_1/theta_2 + id_2/theta_1 = (-1/7)/(-8/28) + (-2/8)/(-7/42) = 2.
%e A336013 [2, 8, 28] + [3, 18, 102] = [5, 26, 130] because
%e A336013 id_1/theta_2 + id_2/theta_1 = (-28/8)/(-18/3) + (-102/18)/(-8/2) = 2.
%e A336013 [28, 8, 2] + [102, 18, 3] = [130, 26, 5] because
%e A336013 id_1/theta_2 + id_2/theta_1 = (-2/8)/(-18/102) + (-3/18)/(-8/28) = 2.
%e A336013 Examples of three rows that sum to a row:
%e A336013 [1, 2, 2] + [1, 2, 2] + [1, 5, 20] = [3, 9, 24] because
%e A336013 (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.
%e A336013 [1, 7, 42] + [2, 8, 28] + [3, 10, 30] = [6, 25, 100] because
%e A336013 (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.
%e A336013 For n=4,
%e A336013 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,
%e A336013 then [x1+x2+x3+x4 , y1+y2+y3+y4 , z1+z2+z3+z4] is a row.
%e A336013 [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
%o A336013 (PARI) for (y = 2, 8, fordiv (y^2-y, x, print([x, y, (y^2-y)/x]) ) ) \\ _David Lovler_, Mar 12 2021
%Y A336013 The number of rows for each y beginning with y=2 is A092517.
%Y A336013 Cf. A332083.
%K A336013 nonn,tabf
%O A336013 1,2
%A A336013 _David Lovler_, Jul 07 2020
%E A336013 Table format in example edited by _David Lovler_, Feb 14 2022