%I #41 Nov 26 2025 15:59:40

%S 0,4,0,18,32,0,42,44,48,24,42,12,40,8,50,368,16,100,410,118,0,12,442,

%T 584,546,1104,482,148,2786,536,398

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

%C 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 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 This is an excerpt which shows the smallest solutions with up to 50 digits only:

%C .n..#digits.....................................smallest.solution......ref

%C .2........4..................................................2178..A072141

%C .4.......18....................................169140971830859028..A292634

%C .5.......32......................10591266563195008940873343680499..A292635

%C .7.......42............142710354353443018141857289645646556981858..A292856

%C .8.......44..........16914079504181797053273763831171860502859028..A292857

%C .9.......48......111603518721165960373027269626940447783074704878..A292858

%C 10.......24..............................101451293600894707746789..A292859

%C 11.......42............166425621223026859056339052269787863565428..A292846

%C 12.......12..........................................118722683079..A072718

%C 13.......40..............1195005230033599502088049947699664004979..A292992

%C 14........8..............................................11436678..A072142

%C 15.......50....10695314508256806604321090888649339244708568530399..A292993

%C 17.......16......................................1186781188132188..A072719

%C 22.......12..........................................108811891188..A072143

%C 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 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 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 J. H. E. Cohn, <a href="https://www.fq.math.ca/Scanned/28-2/cohn.pdf">Palindromic Differences</a>, Fibonacci Quart. (1990):113-120.

%H Thorsten Ehlers, <a href="/A215669/a215669.txt">smallest solutions for the reverse-and-subtract problem up to cycle length 31</a>

%e a(4) = 169140971830859028 as the smallest cycle with length 4 is 169140971830859028 -> 651817066348182933 -> 312535222687464777 -> 464929563535070436 ( -> 169140971830859028 ).

%Y Cf. A072141, A072142, A072143, A072718, A072719, A292634, A292635, A292846, A292856, A292857, A292858, A292859, A292992, A292993.

%K base,nonn,more

%O 1,2

%A _Thorsten Ehlers_, Aug 20 2012

%E Added a reference, formatted and added one more example in comments. - _Thorsten Ehlers_, Oct 06 2012

%E Sequences added to comments and crossrefs by _Ray Chandler_, Sep 27 2017