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 A215669 Number of decimal digits of the smallest solution for the reverse-and-subtract problem for cycle length n. 10
 0, 4, 0, 18, 32, 0, 42, 44, 48, 24, 42, 12, 40, 8, 50, 368, 16, 100, 410, 118, 0, 12, 442, 584, 546, 1104, 482, 148, 2786, 536, 398 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Solution x for a given cycle length n for the reverse-and-subtract problem is defined as x = f^n(x), x <> f^j(x) for j < n, where f: k -> |k - reverse(k)|. For some cycle lengths (at least for 1, 3, 6 and 21) no solutions exist, these are marked as 0 in above sequence. Zero cannot be considered a solution for cycle length 1 as there are nontrivial solutions for other numeral systems, such as 13 (one-three) in base 5 numeral system. This is an excerpt which shows the smallest solutions with up to 50 digits only: .n..#digits.....................................smallest.solution......ref .2........4..................................................2178..A072141 .4.......18....................................169140971830859028..A292634 .5.......32......................10591266563195008940873343680499..A292635 .7.......42............142710354353443018141857289645646556981858..A292856 .8.......44..........16914079504181797053273763831171860502859028..A292857 .9.......48......111603518721165960373027269626940447783074704878..A292858 10.......24..............................101451293600894707746789..A292859 11.......42............166425621223026859056339052269787863565428..A292846 12.......12..........................................118722683079..A072718 13.......40..............1195005230033599502088049947699664004979..A292992 14........8..............................................11436678..A072142 15.......50....10695314508256806604321090888649339244708568530399..A292993 17.......16......................................1186781188132188..A072719 22.......12..........................................108811891188..A072143 Solutions for all cycle lengths up to 31 can be found below in the links section. Remember that a zero means there exists no solution for this specific cycle length. There are two ways to find such solutions, first you can search in a given range of numbers e.g. from 10000000 to 99999999 and apply reverse-and-subtract to each number until you fall below the smallest number in this range (here: 10000000) or you find a cycle. Obviously, this works well only on small numbers up to 18-20 digits. The second way is to construct a cycle with a given length n from the outside in until the innermost 2 digits of each number match the conditions for a valid cycle. This way it is possible to get the above results within seconds up to some hours depending on the specific cycle length even on an outdated PC. LINKS Table of n, a(n) for n=1..31. J. H. E. Cohn, Palindromic Differences, Fibonacci Quart. (1990):113-120. Thorsten Ehlers, smallest solutions for the reverse-and-subtract problem up to cycle length 31 EXAMPLE a(4) = 169140971830859028 as the smallest cycle with length 4 is 169140971830859028 -> 651817066348182933 -> 312535222687464777 -> 464929563535070436 ( -> 169140971830859028 ). CROSSREFS Cf. A072141, A072142, A072143, A072718, A072719, A292634, A292635, A292846, A292856, A292857, A292858, A292859, A292992, A292993. Sequence in context: A233807 A302771 A167350 * A244310 A334705 A156457 Adjacent sequences: A215666 A215667 A215668 * A215670 A215671 A215672 KEYWORD base,nonn,more AUTHOR Thorsten Ehlers, Aug 20 2012 EXTENSIONS Added a reference, formatted and added one more example in comments. - Thorsten Ehlers, Oct 06 2012 Sequences added to comments and crossrefs by Ray Chandler, Sep 27 2017 STATUS approved

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Last modified August 11 12:11 EDT 2024. Contains 375069 sequences. (Running on oeis4.)