cp's OEIS Frontend

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is possible to get the above results within seconds up to some hours depending on the specific cycle length even on an outdated PC.

There are 15 fifty-digit terms in the sequence. Further terms are obtained (a) by inserting at the center of these terms either any number of 0's (for 10695314508256806604321090888649339244708568530399, 26730889210860738952361172793674105293199801097128, 29899105876561459824028272726867015583422139910097, 49102887245877091252834454555175879833145710289795, 55448121688278511195278554322651878413601497706634, 68315444154984874470735536347381553142144945548514, 88608272072487486790367718123691321620571972829202) or any number of 9's (for the other eight terms) and (b) by concatenating a term any number of times with itself and inserting an equal number of 0's at all junctures. Method (b) may be applied recursively to all terms. - Ray Chandler, Oct 15 2017