%I A107437
%S A107437 5,7,5,5,5,5,6,5,5,5,6,5,6,5,5,5,5,6,7,5,5,6,5,7,5,5,5,6,5,5,5,5,5,5,5,
%T A107437 5,5,5,5,5,5,5,5,5,5,5,6,5,6,5,5,5,7,5,5,5,5,6,5,8,5,6,5,6,5,7,5,5,6,6,
%U A107437 5,7,5,5,6,5,5,6,5,5,5,5,5,5,6,5,7,5,6,5,6,5,5,7,9,5,5,5,7,5,7,5,5,8,5
%N A107437 Number of difference boxes for initial a=prime(n), b=prime(n+1), c=prime(n+1), 
               s=prime(n+3) (see comments).
%C A107437 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.
%D A107437 A. Behn et al., The convergence of difference boxes, Amer. Math. Month. 
               v. 112 (2005), pp. 426-439.
%t A107437 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
%Y A107437 Sequence in context: A165242 A104542 A161376 this_sequence A122271 A011205 
               A167133
%Y A107437 Adjacent sequences: A107434 A107435 A107436 this_sequence A107438 A107439 
               A107440
%K A107437 easy,nonn
%O A107437 1,1
%A A107437 Zak Seidov (zakseidov(AT)yahoo.com), Jun 09 2005