Solution x for a given cycle length n for the reverseandsubtract 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 (onethree) 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 reverseandsubtract 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 1820 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is possible to get the above results within seconds up to some hours depending on the specific cycle length even on an outdated PC.
