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%I #12 Jul 25 2017 06:03:02
%S 99,891,198,792,297,693,396,594,495
%N If a1a2a3 is a 3-digit integer in a concatenated form, we define two permutations of its digits as follows: P1(a1a2a3)=a2a3a1 and P2(a1a2a3)=a1a3a2, then we take the absolute value of their difference. Thus we form a sequence: a1a2a3, abs(P1(a1a2a3)-P2(a1a2a3)), and so on.
%C This is an alternative to Kaprekar's routine. It would be interested in studying 4-digit integers with the permutations P1(a1a2a3a4)=a2a3a4a1 and P2(a1a2a3a4)=a1a3a4a2. Other permutations can be also studied. A generalization of Kaprekar's routine is the following: Let f be an operator that maps a finite set A={a1, a2, ..., a_p}, with p>=1 elements, into itself. Then, for any value 'a' in A, we have f(a) belongs to A too. If we iterate this operator multiple times, we get a chain: a, f(a), f(f(a)), ..., f(f...f(a)...), ... all of whose elements are in A. But, since A is finite, after at most p iterations we get two equal iterations. Therefore we end up in a finite cycle (of one or more terms).
%H F. Smarandache, <a href="http://fs.gallup.unm.edu//ProposedProblems2.pdf">Proposed Problems of Mathematics, Vol. II</a>, State University of Moldova Press, Kishinev, pp. 83-84, 1997.
%H F. Smarandache, <a href="http://fs.gallup.unm.edu//MultispaceMultistructure.pdf">Generalization and alternatives of Kaprekar's routine</a>, Multispace & Multistructure / Neutrosophic Transdisciplinarity, Northern-European Publishers, pp. 555-559, Finland, 2010. <a href="http://arxiv.org/abs/1005.3235">arXiv:1005.3235</a>
%F abs(P1(a1a2a3)-P2(a1a2a3)) = abs(a2a3a1-a1a3a2) = 99x(a2-a1).
%e Starting with 100, we get abs(001-100)=099, then abs(990-099)=891, then abs(918-819)=099, etc. So 100, 099, 891, 099, ... (the cycle is 099, 891). Each three-digit number ends up in a cycle of two terms (such as: 99 and 891, or 198 and 792, or 297 and 693, or 396 and 594), or in a constant 495 (as in Kaprekar's routine).
%e Starting with 495, we get abs(954-459)=495 (cycle of one term).
%Y Cf. A090429, A099010.
%K nonn,base,fini,full,less
%O 1,1
%A F. Smarandache (smarand(AT)unm.edu), May 10 2010
%E Added keyword:base,fini,full as there are only 9 different values obtained by the abs() starting from any a1a2a3 in the range 100 to 999 _R. J. Mathar_, May 15 2010
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