A364057 Lexicographically earliest infinite sequence of positive integers such that every subsequence {a(j), a(j+k), a(j+2k)} (j, k >= 1) is unique.
1, 1, 1, 2, 3, 1, 2, 4, 5, 6, 7, 8, 5, 1, 2, 9, 3, 4, 6, 7, 1, 10, 11, 12, 13, 8, 14, 15, 16, 3, 17, 9, 18, 4, 7, 19, 5, 2, 11, 12, 20, 6, 1, 8, 21, 22, 9, 23, 24, 13, 14, 3, 10, 16, 17, 25, 26, 19, 27, 6, 28, 11, 15, 20, 22, 29, 12, 21, 16, 23, 30, 18, 31, 32
Offset: 1
Keywords
Examples
For a(9), we first try 1. If a(9) were 1, {a(3), a(6), a(9)} would be {1, 1, 1}, but this already occurred at {a(1), a(2), a(3)}. Next, try 2. If a(9) were 2, {a(3), a(6), a(9)} would be {1, 1, 2}, but this already occurred at {a(2), a(3), a(4)}. Next, try 3. If a(9) were 3, {a(3), a(6), a(9)} would be {1, 1, 3}, but this already occurred at {a(1), a(3), a(5)}. Next, try 4. If a(9) were 4, {a(1), a(5), a(9)} would be {1, 3, 4}, but this already occurred at {a(2), a(5), a(8)}. Then, try 5. New subsequences at indices {a(1), a(5), a(9)} = {1, 3, 5}, {a(3), a(6), a(9)} = {1, 1, 5}, {a(5), a(7), a(9)} = {3, 2, 5}, and {a(7), a(8), a(9)} = {2, 4, 5} are formed, none of which have occurred at any {a(j), a(j+k), a(j+2k)} (for any j and k) previously. No 5 has occurred previously, so criteria (2) in Comments must be satisfied. Thus a(9) = 5. a(10) is the first time a candidate is denied solely because it would create a guaranteed future duplicate. Note that no subsequences prevent a(10) from being 4. n = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 a(n) = 1 1 1 2 3 1 2 4 5 [4] X | | | | | | If a(10) were 4, {a(2), a(8), a(14)} = {a(6), a(10), a(14)} = {1, 4, X}, making a subsequence {a(j), a(j+k), a(j+2k)} which is not unique. Therefore a(10) != 4.
Links
- Rémy Sigrist, Table of n, a(n) for n = 1..10000
- Samuel Harkness, MATLAB program
- Rémy Sigrist, C++ program
Programs
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MATLAB
See Links section. (C++) See Links section.
Comments