1,1

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 (more entries needed to submit to OEIS!) and there are many cases with k=10. Also, it is evident expansion for polygons, someone may wish to consider it.

A. Behn et al., The convergence of difference boxes, Amer. Math. Month. v. 112 (2005), pp. 426-439.

s={}; Do[{aa=Prime[n], bb=Prime[n+1], cc=Prime[n+2], dd=Prime[n+3]}; Do[a=Abs[bb-aa]; b=Abs[cc-bb]; c=Abs[dd-cc]; d=Abs[aa-dd]; If[a\[Equal]b\[Equal]c\[Equal]d\[Equal]0, s=Append[s, k]; Break[]]; aa=a; bb=b; cc=c; dd=d, {k, 100}], {n, 1, 200}]; s

Sequence in context: A165242 A104542 A161376 this_sequence A122271 A011205 A167133

Adjacent sequences: A107434 A107435 A107436 this_sequence A107438 A107439 A107440

easy,nonn

Zak Seidov (zakseidov(AT)yahoo.com), Jun 09 2005