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

'Reverse and Subtract' (cf. A070837, A070838) is defined by x -> |x - reverse(x)|, where reverse(x) is the digit reversal of x.

For every n the trajectory eventually becomes periodic, since 'Reverse and Subtract' does not increase the number of digits and so the set of available terms is finite. For small n the period length is 1, the periodic part consists of 0's, the last term of the preperiodic part is a palindrome.

The first n with period length 2 and a nontrivial periodic part is 1012 (cf. A072140).

This sequence is a weak analog of A033665, which uses 'Reverse and Add'.

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 12 twelve-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 178574614218, 277673713317, 614218178574, 713317277673) 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. - Revised thanks to a comment from Hans Havermann, Jan 27 2004.

'Reverse and Subtract' (cf. A072137) is defined by x -> |x - reverse(x)|. There is no number k > 0 such that |k - reverse(k)| = k, so 0 is the only period with length 1. Consequently this sequence consists of the numbers n such that repeated application of 'Reverse and Subtract' does not lead to a palindrome. It is an analog of A023108, which uses 'Reverse and Add'. - A072141, A072142, A072143 give the numbers which generate periods of length 2, 14, 22 respectively.

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.

In the definition, j can be restricted to proper divisors of k. A073143 gives the corresponding values of k. A073144(n) gives the smallest m such that the 'Reverse and Subtract' trajectory of m leads to a(n). Presumably a(6) = 1186781188132188 with k = 17.

Presumably a(6) = 17. a(n) is the length of the periodic part (cf. A072137) of the trajectory of A073142(n). Question: Does every k > 0 appear in this sequence?

Presumably a(6) = 1000000011011012.