A177686 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.

99, 891, 198, 792, 297, 693, 396, 594, 495

Offset: 1

Author

F. Smarandache (smarand(AT)unm.edu), May 10 2010

Keywords

• nonn
• base
• fini
• full
• less

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).

Examples

			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).
		

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

Crossrefs

Cf. A090429, A099010.

Formula

abs(P1(a1a2a3)-P2(a1a2a3)) = abs(a2a3a1-a1a3a2) = 99x(a2-a1).

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