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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

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