OFFSET
0,5
COMMENTS
To calculate a(n) take all the terms from a(0) to a(n-1), concatenate them, and then count the distinct digit strings that have at least two nonoverlapping occurrences. For example, if the concatenated terms formed the string '2210102240404' then the next term would be 8 as the strings '0','1','2','4','04','10','22','40' have all occurred at least twice. Note that the string '404' is not counted as its two occurrences overlap.
In the first 20000 terms, the largest increase in consecutive terms is from a(16496) = 45375 to a(16497) = 45410, an increase of 35; the concatenation of a(16496) to the previous terms results in the longest repeated string, '245352453624537'.
EXAMPLE
a(1) = 0 as there have been no repeated strings prior to a(1).
a(2) = 1 as there has been one repeat of '0', which occurs in a(0) and a(1).
a(3) = 1 as only '0' has repeated up to a(2).
a(4) = 2 as now both '0' and '1' have repeated up to a(3).
a(22) = 12 as the strings '0' to '10' have repeated, but '10,10' contains the string '01' which also appears from a(1) to a(2), thus '01' has also repeated.
a(23) = 13 as a(22) = '12' created a repeat of the string '12', with a(3) and a(4). Even though the terms '10,10,12' contain two occurrences of the string '101' they overlap so are not counted. The string '10,12' also contains another '01' but that has already repeated and been counted previously so is ignored as the count is of distinct strings.
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Scott R. Shannon, Nov 28 2019
STATUS
approved