cp's OEIS Frontend

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).

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?