A107437 - OEIS

Put any four numbers, a, b, c, d in the vertices of the square; in the middle of each side write unsigned difference between numbers at the side's ends and get another four numbers a, b, c, d in the vertices of the smaller square. Repeat the procedure until a=b=c=d=0. It's evident that for any initial a, b, c, d, the final numbers are all zero, simply because each of a, b, c, d only decreases at each step. But it seems (see Reference) that it is proved (or only suggested?) to be true also for signed differences. In some particular cases it is possible to calculate exactly all steps For example, let a=n; b=n^2; c=n^3; d=n^4; then we have subsequently: k=0: a=n, b=n^2, c=n^3, d=n^4; k=1: a=n( n-1), b=(n-1)*n^2, c=(n-1)*n^3, d=n(n^3-1); k=2: a=n(n-1)^2, b=(n-1)^2*n^2, c=n(n^2-1), d=n^2*(n^2-1); k=3: a=n(n-1)^3, b= n(1+n -3n^2+n^3), c=n(n+1)(n-1)^2, d=n(n-1)(1+n^2 ); k=4: a=2n(n-1), b=2(n-1)n^2, c=a, d=b; k=5: a=b=c=d=2n(n-1)^2; k=6: a=b=c=d=0. We say that in k=6 steps we come to finish, at any n. The sequence gives the number of k(n) for the case k=0: a=Prime[n], b=Prime[n+1], c=Prime[n+2], d=Prime[n+3]}. Maximal found k is 12 at n=22059; three cases with k=11 are at n=18024, 41761, 84938 and there are many cases with k=10. Also, it is evident expansion for polygons, someone may wish to consider it.