OFFSET
1,1
COMMENTS
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).
LINKS
Ray Chandler, Table of n, a(n) for n = 1..10000
Michael P. Greaney, 1012 and other such numbers, +plus magazine, August 30, 2017.
FORMULA
n = f(f(n)), n <> f(n), where f: x -> |x - reverse(x)|.
EXAMPLE
6534 -> |6534 - 4356| = 2178 -> |2178 - 8712| = 6534.
CROSSREFS
KEYWORD
base,nonn
AUTHOR
Klaus Brockhaus, Jun 24 2002
EXTENSIONS
More terms from Ray Chandler, Oct 09 2017
STATUS
approved