A367849 Lexicographically earliest infinite sequence of positive integers such that for each n, the values in a path of locations starting from any i=n are all distinct, where jumps are allowed from location i to i+a(i).

1, 1, 2, 1, 3, 1, 1, 4, 1, 1, 2, 5, 3, 1, 1, 4, 6, 1, 1, 5, 1, 1, 7, 1, 6, 1, 1, 2, 1, 8, 7, 1, 1, 2, 1, 3, 1, 9, 4, 1, 1, 2, 5, 3, 1, 1, 10, 6, 1, 1, 2, 1, 3, 7, 1, 4, 11, 1, 1, 5, 8, 1, 1, 2, 6, 3, 1, 12, 9, 1, 7, 1, 1, 2, 1, 3, 1, 10, 4, 13, 1, 1, 5, 1, 1, 2, 1, 11, 1, 1, 2, 1, 14, 1, 1, 2, 1, 3

Offset: 1

Author

Neal Gersh Tolunsky, Dec 08 2023

Keywords

• nonn

Comments

Consider each index i as a location from which one can jump a(i) terms forward. No starting index can reach the same value more than once by forward jumps.

The value a(i) at the starting index is not part of the path (and thus allows a(2)=1).

The indices of first occurrences are given by A133263 (essentially triangular numbers + 2).

Changing the definition so that jumps are allowed only from location i to i-a(i) gives A002260.

Examples

			We can see, for example, that the terms reachable by jumping forward continuously from i=1 are all distinct (and in this case are just the positive integers):
  1, 1, 2, 1, 3, 1, 1, 4, 1, 1, 2, 5
  *->1->2---->3------->4---------->5
Beginning at i=9 and jumping forward continuously, we get the sequence 1,2,3,4,5,6,7,9 (in which all terms are likewise distinct).
		

Links

• Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000
• Thomas Scheuerle, Plot of a(1..100000)^2. It is conjectured that only the topmost straight line in this plot will extend into infinity.
• Thomas Scheuerle, Partial view of a(1..80000)^2 equalized to Y axis by shear mapping. This shows the different lengths of the increasing sequences.
• Neal Gersh Tolunsky, Ordinal transform of 20000 terms
• Neal Gersh Tolunsky, First differences of 20000 terms

Crossrefs

Cf. A367467, A367832, A133263 (index of first occurrences), A362248.

Programs

MATLAB

function a = A367849( max_n )
    a = zeros(1,max_n); j = find(a == 0,1);
    while ~isempty(j)
        a(j) = 1; k = 1;
        if j+k < max_n
            while a(j+k) == 0
                a(j+k) = k;
                if j+2*k-1 < max_n
                    j = j+(k-1); k = k+1;
                else
                    break;
                end
            end
        end
        j = find(a == 0,1);
    end
end % Thomas Scheuerle, Dec 09 2023

Formula

a(A133263(n)) = n + 1.