OFFSET
1,1
COMMENTS
This sequence is aperiodic. Proof: If any number of initial terms are removed, said initial terms can be recreated by extrapolating backwards from later terms. In other words, the later terms include a perfect description of the earlier terms. This implies that, if the sequence is periodic, the repeating section must include all of the initial terms of the sequence in the order in which they originally appear. This is impossible because we cannot extrapolate backwards infinitely from the initial terms. If we attempt to do so, we get the pattern 5, 5, 4, 2, 4, 5 (the last three values being the first three terms of the sequence). This pattern is clearly invalid, so the sequence must be aperiodic. - Samuel B. Reid, Jun 07 2020
LINKS
Samuel B. Reid, Table of n, a(n) for n = 1..10000
Samuel B. Reid, Python program for A309681
Samuel B. Reid, C program for A309681
EXAMPLE
If a(2) were 1, a(2) would have to be 2. a(2) cannot be both 1 and 2, so a(2) cannot be 1.
If a(2) were 2, the next 2 after a(1) would appear at a(3). The next 2 actually appears at a(2), so a(2) cannot be 2.
If a(2) were 3, a(4) would have to be 2. If a(4) were 2, a(3) and a(5) would have to have the same value. If a(3) and a(5) were equal, the pattern X, 3, Y, 2 would appear later in the sequence. The pattern X, 3, Y, 2 is invalid because X would have to be 2. X cannot be 2 because it comes after the 2 at a(1) and before the 2 that succeeds Y. This behavior is inconsistent with the definition of the sequence.
The next possible value for a(2) is 4, which causes no trivial inconsistencies (as with a(2)=1 or a(2)=2) or logical inconsistencies (as with a(2)=3).
The n-th term of the sequence is the lowest value that does not cause inconsistencies of either sort.
PROG
(Python and C) See Links section.
CROSSREFS
KEYWORD
nonn
AUTHOR
Samuel B. Reid, Aug 12 2019
STATUS
approved