A351581 - OEIS

%I #29 Mar 17 2022 23:58:04

%S 2,2,2,1,3,3,3,2,12,6,3,1,1,2,4,6,4,4,4,3,9,6,4,2,36,12,4,1,5,5,5,4,

%T 20,10,5,2,80,20,5,1,6,6,6,5,150,30,6,1,7,7,7,6,28,14,7,3,63,21,7,2,

%U 252,42,7,1,2,4,8,14,8,8,8,7,98,28,8,2,392,56,8,1

%N Four-column table read by rows giving quadruples of integers [w,x,y,z] with w > 0, x > 1, y > 1 and z > 0 such that y^2 - y - x*z = 0 and x^2 = w*y, sorted by y then by x.

%C It is the same to sort by y then by w also to sort by y then by z descending.

%C When [w,x,y,z] is a row, f(a,b,c) = w*a*b*c + x*(a*b + a*c + b*c) + y*(a+b+c) + z is associative in the following sense. f((a,b,c),d,e) = f(a,f(b,c,d),e) = f(a,b,f(c,d,e)) for all a,b,c,d,e. f(a,b,c) is commutative because of its symmetry.

%C For each quadruple, the corresponding f(a,b,c) has a unique zero element (call it theta), meaning f(a,b,theta) = f(a,theta,b) = f(theta,a,b) = theta for all a,b. Theta = -y/x = - x/w. f(a,b,c) also has not one but two identity elements (id_1 and id_2), meaning f(a,id_1,id_1) = f(id_1,a,id_1 = f(id_1,id_1,a) = a for all a and f(a,id_2,id_2) = f(id_2,a,id_2) = f(id_2,id_2,a) = a for all a. Id = (-y +- sqrt(y))/x = theta +- sqrt(y)/x. Thus theta = (id_1 + id_2)/2.

%C The identity elements are integers when y is a square and x divides sqrt(y).

%H David Lovler, <a href="/A351581/b351581.txt">Table of n, a(n) for n = 1..2720</a>

%H David Lovler, <a href="/A351581/a351581.txt">The first 680 quadruples for y up to 100</a>.

%F Looking at A336013, if [X,Y,Z] is a row and f(a,b) = X*a*b + Y*(a+b) + Z is the corresponding associative function with id = -Z/Y and theta = -Y/X, then the composition f(f(a,b),c) = w*a*b*c + x*(a*b + a*c + b*c) + y*(a+b+c) + z = f(a,b,c) gives the quadruple [w,x,y,z]. f(a,b,c) has the same theta as f(a,b); the two identity elements for f(a,b,c) are id and 2*theta - id.

%F If theta and the identity elements are computed from a quadruple, f(a,b,c) can be written as (a*b*c - theta*(a*b + a*c + b*c) + theta^2*(a+b+c) - theta^3)/(id-theta)^2 + theta. The square in the denominator ensures that f(a,b,c) is the same for either id.

%F Two parameters are sufficient to describe a row. For n*s > 1, rows are [w,x,y,z] = [n, n*s, n*s^2, (n^2*s^4-n*s^2)/(n*s)] = [n, n*s, n*s^2, n*s^3 - s]. In terms of n and s, theta = -s and id = s*(-1 +- 1/sqrt(n)). Rows with s=1 stand out as having w=x=y; theta = -1 and id = -1 +- 1/sqrt(w).

%e Table begins:

%e   [ w,  x,  y,  z]

%e   -----------------

%e   [  2,  2, 2,  1];

%e   [  3,  3, 3,  2];

%e   [ 12,  6, 3,  1];

%e   [  1,  2, 4,  6];

%e   [  4,  4, 4,  3];

%e   [  9,  6, 4,  2];

%e   [ 36, 12, 4,  1];

%e   [  5,  5, 5,  4];

%e   [ 20, 10, 5,  2];

%e   [ 80, 20, 5,  1];

%e   [  6,  6, 6,  5];

%e   [150, 30, 6,  1];

%e   [  7,  7, 7,  6];

%e   [ 28, 14, 7,  3];

%e   [ 63, 21, 7,  2];

%e   [252, 42, 7,  1];

%e   [  2,  4, 8, 14];

%e   [  8,  8, 8,  7];

%e   [ 98, 28, 8,  2];

%e   [392, 56, 8,  1];

%e   [  1,  3, 9, 24];

%e   [  4,  6, 9, 12];

%e   [  9,  9, 9,  8];

%e   [ 16, 12, 9,  6];

%e   [ 36, 18, 9,  4];

%e   [ 64, 24, 9,  3];

%e   [144, 36, 9,  2];

%e   [576, 72, 9,  1];

%e   ...

%e For row [1, 2, 4, 6], f(a,b,c) = a*b*c + 2*(a*b + a*c + b*c) + 4*(a+b+c) + 6. Theta = -2; id_1 = -1, id_2 = -3. The associative function f(a,b) = a*b + 2*(a+b) + 2 has theta = -2 and id = -1; f(f(a,b),c) = f(a,b,c). Another associative function g(a,b) = -a*b - 2*(a+b) - 6 with theta = -2 and id = -3 likewise gives g(g(a,b),c) = f(a,b,c).

%o (PARI) { my(y); for (y = 2, 9, fordiv (y^2-y, x, if(type(w = x^2/y) == "t_INT", print([w, x, y, (y^2-y)/x]) )) ) }

%Y Cf. A336013, A351986.

%K nonn,tabf

%O 1,1

%A _David Lovler_, Feb 13 2022