OFFSET
1,1
COMMENTS
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).
LINKS
F. Smarandache, Proposed Problems of Mathematics, Vol. II, State University of Moldova Press, Kishinev, pp. 83-84, 1997.
F. Smarandache, Generalization and alternatives of Kaprekar's routine, Multispace & Multistructure / Neutrosophic Transdisciplinarity, Northern-European Publishers, pp. 555-559, Finland, 2010. arXiv:1005.3235
FORMULA
abs(P1(a1a2a3)-P2(a1a2a3)) = abs(a2a3a1-a1a3a2) = 99x(a2-a1).
EXAMPLE
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).
Starting with 495, we get abs(954-459)=495 (cycle of one term).
CROSSREFS
KEYWORD
nonn,base,fini,full,less
AUTHOR
F. Smarandache (smarand(AT)unm.edu), May 10 2010
EXTENSIONS
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
STATUS
approved