cp's OEIS Frontend

There are 14 eight-digit terms in the sequence. Further terms are obtained (a) by inserting at the center of these terms either any number of 9's (for 11436678, 32662377, 33218856, 76226733) or any number of 0's (for the other ten 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. - Revised thanks to a comment from Hans Havermann, Jan 20 2004.

There are 22 twelve-digit terms in the sequence. Further terms are obtained (a) by inserting at the center of these terms any number of 9's 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. - Revised thanks to a comment from Hans Havermann, Jan 27 2004

All terms are divisible by 10999989. - Hugo Pfoertner, Sep 23 2020

There are 17 sixteen-digit terms in the sequence. Further terms are obtained (a) by inserting at the center of these terms any number of 9's 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. - Revised thanks to a comment from Hans Havermann, Jan 27 2004.

This is a working sequence. It is neither by computation nor by proof guaranteed that there are no smaller or interleaved terms.

All terms are divisible by 1099999989. - Hugo Pfoertner, Sep 23 2020

There are two four-digit terms in the sequence. Further terms are obtained (a) by inserting at the center of these terms any number of 9's 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. - Revised thanks to a comment from Hans Havermann, Jan 27 2004.

Solutions to x = f^k(x), x <> f^j(x) for j < k, where f: n -> |n - reverse(n)|, for period lengths k <= 22 are given by:

.k..smallest.solution..smallest.n.with.period.k..sequence

.1..................0.........................0.......---

.2...............2178......................1012..(this one)

14...........11436678..................10001145...A072142

22.......108811891188..............100000114412...A072143

12.......118722683079..............100010505595...A072718

17...1186781188132188..........1000000011011012...A072719

I still have no answer to the question if there exist solutions for other values of k. Random tests for larger n (up to 50 digits) have shown that periods 1 and 2 are very frequent (> 90 %), period 14 is not unusual (7 to 8 %), periods 22, 12 and 17 are very rare and other periods did not appear.

I conjecture that for some k there are no solutions, while in other cases the minimal solutions will have 20, 24, 28, ... digits (which however are very hard to find).

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 11 forty-two-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 166425621223026859056339052269787863565428, 311906350108036145286307567270199935391877, 466287189883036620417374974360601118217236, 658139747564935391877311906350534262959233, 703288139752915027377325180481642968027593) or any number of 9's (for the other six 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. - Clarified by Ray Chandler, Oct 14 2017.

There are 7 forty-two-digit terms in the sequence. Terms of derived sequences can be obtained either by inserting at the center of their digits any number of 9's or by concatenating a term any number of times with itself and inserting an equal number of 0's at all junctures.

There are 8 forty-four-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 16914079504181797053273763831171860502859028, 46492964703403651468863122584458570244070436, 65181741002635316783463471248546280094182933) or any number of 9's (for the other five 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. - Clarified by Ray Chandler, Oct 14 2017.

There are 9 forty-eight-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 111603518721165960373027269626940447783074704878, 176512193475025275151977319848516480415708873428, 637711546692957642526533655473763239712353991164, 647866614039059340696936459303056040158682342243, 766803951666578089253935450746129113344740601233) or any number of 9's (for the other four 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. - Clarified by Ray Chandler, Oct 14 2017.

There are 10 twenty-four-digit terms in the sequence. Terms of derived sequences can be obtained either by inserting at the center of their digits any number of 9's or by concatenating a term any number of times with itself and inserting an equal number of 0's at all junctures.