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A215669 Number of decimal digits of the smallest solution for the reverse-and-subtract problem for cycle length n.

%I A215669 #40 Jan 05 2025 19:51:39
%S A215669 0,4,0,18,32,0,42,44,48,24,42,12,40,8,50,368,16,100,410,118,0,12,442,
%T A215669 584,546,1104,482,148,2786,536,398
%N A215669 Number of decimal digits of the smallest solution for the reverse-and-subtract problem for cycle length n.
%C A215669 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.
%C A215669 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.
%C A215669 This is an excerpt which shows the smallest solutions with up to 50 digits only:
%C A215669 .n..#digits.....................................smallest.solution......ref
%C A215669 .2........4..................................................2178..A072141
%C A215669 .4.......18....................................169140971830859028..A292634
%C A215669 .5.......32......................10591266563195008940873343680499..A292635
%C A215669 .7.......42............142710354353443018141857289645646556981858..A292856
%C A215669 .8.......44..........16914079504181797053273763831171860502859028..A292857
%C A215669 .9.......48......111603518721165960373027269626940447783074704878..A292858
%C A215669 10.......24..............................101451293600894707746789..A292859
%C A215669 11.......42............166425621223026859056339052269787863565428..A292846
%C A215669 12.......12..........................................118722683079..A072718
%C A215669 13.......40..............1195005230033599502088049947699664004979..A292992
%C A215669 14........8..............................................11436678..A072142
%C A215669 15.......50....10695314508256806604321090888649339244708568530399..A292993
%C A215669 17.......16......................................1186781188132188..A072719
%C A215669 22.......12..........................................108811891188..A072143
%C A215669 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.
%C A215669 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.
%C A215669 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.
%H A215669 J. H. E. Cohn, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/28-2/cohn.pdf">Palindromic Differences</a>, Fibonacci Quart. (1990):113-120.
%H A215669 Thorsten Ehlers, <a href="/A215669/a215669.txt">smallest solutions for the reverse-and-subtract problem up to cycle length 31</a>
%e A215669 a(4) = 169140971830859028 as the smallest cycle with length 4 is 169140971830859028 -> 651817066348182933 -> 312535222687464777 -> 464929563535070436 ( -> 169140971830859028 ).
%Y A215669 Cf. A072141, A072142, A072143, A072718, A072719, A292634, A292635, A292846, A292856, A292857, A292858, A292859, A292992, A292993.
%K A215669 base,nonn,more
%O A215669 1,2
%A A215669 _Thorsten Ehlers_, Aug 20 2012
%E A215669 Added a reference, formatted and added one more example in comments. - _Thorsten Ehlers_, Oct 06 2012
%E A215669 Sequences added to comments and crossrefs by _Ray Chandler_, Sep 27 2017