

A215669


Number of decimal digits of the smallest solution for the reverseandsubtract problem for cycle length n.


10



0, 4, 0, 18, 32, 0, 42, 44, 48, 24, 42, 12, 40, 8, 50, 368, 16, 100, 410, 118, 0, 12, 442, 584, 546, 1104, 482, 148, 2786, 536, 398
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OFFSET

1,2


COMMENTS

Solution x for a given cycle length n for the reverseandsubtract 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 (onethree) 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 reverseandsubtract 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 1820 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.


LINKS



EXAMPLE

a(4) = 169140971830859028 as the smallest cycle with length 4 is 169140971830859028 > 651817066348182933 > 312535222687464777 > 464929563535070436 ( > 169140971830859028 ).


CROSSREFS

Cf. A072141, A072142, A072143, A072718, A072719, A292634, A292635, A292846, A292856, A292857, A292858, A292859, A292992, A292993.


KEYWORD

base,nonn,more


AUTHOR



EXTENSIONS

Added a reference, formatted and added one more example in comments.  Thorsten Ehlers, Oct 06 2012
Sequences added to comments and crossrefs by Ray Chandler, Sep 27 2017


STATUS

approved



