A367467 Lexicographically earliest infinite sequence of positive integers such that a(n + a(n)) is distinct for all n.

1, 1, 2, 2, 3, 4, 2, 5, 6, 7, 1, 8, 9, 2, 10, 11, 12, 1, 13, 14, 2, 15, 16, 2, 17, 18, 19, 2, 20, 21, 2, 22, 23, 24, 1, 25, 26, 2, 27, 28, 2, 29, 30, 31, 2, 32, 33, 2, 34, 35, 36, 1, 37, 38, 2, 39, 40, 41, 1, 42, 43, 2, 44, 45, 2, 46, 47, 48, 1, 49, 50, 2, 51, 52, 53, 1, 54, 55, 2, 56, 57, 2, 58, 59, 60, 2, 61, 62, 2

Offset: 1

Author

Neal Gersh Tolunsky, Nov 18 2023

Keywords

• nonn

Comments

Consider each index i as a location from which one can jump a(i) terms forward. To find a(n) we have to check 2 conditions:

1. The value a(n) can be reached in one jump by at most one previous location.

2. Location n reaches a location in one jump that is not reached in one jump from a location before n.

Described in the above way, the sequence seems to be structured as follows:

A083051 appears to give the indices which cannot be reached from any earlier term; the terms at these indices are 1s and 2s.

A087057 appears to give the indices which can be reached from an earlier term; except for a(2), these terms are first occurrences.

From Thomas Scheuerle, Nov 26 2023: (Start)

Empirical observations:

It appears that this sequence consists of the natural numbers in ascending order interspersed by 1 and 2.

If we consider the distance between successive ones, we will observe a nonperiodic pattern: 9,7,17,17,7,10,7,17,7,10,... . It appears that there are only 7, 10 and 17 with the exception of 9 once.

If we consider the distance between successive twos, we will also observe an interesting nonperiodic pattern: 3,7,7,3,4,3,7,3,4,3,7,7,3,... . It appears that this pattern consists only of 3, 4 and 7. (End)

Examples

			Initial locations and the (by definition) distinct terms that they reach:
     n|  1  2  3  4  5  6  7  8  9
  a(n)|  1  1  2  2  3  4  2  5  6
          =>1=>2====>3
                   ====>4
                      =======>5
                            ====>6
When we evaluate a(i+a(i)) with each index i, we get a distinct value. When i=1, for example, a(1+a(1))=a(1+1)=a(2)=1;  no other i gives 1 as the solution to a(i+a(i)). When i=4, a(4+a(4))=a(4+2)=a(6)=4, and 4 is likewise a solution unique to i=4.
		

Links

• Neal Gersh Tolunsky, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A367832, A293078, A323420, A359807, A367039, A337226, A366691.

Cf. A049472, A083051, A087057, A184119, A328987.

Programs

MATLAB

function a = A367467( max_n )
    a = [1 1:2*max_n];
    for n = 3:max_n
        a(n) = 1;
        while consistency(a, n) == false
            a(n) = a(n)+1;
        end
    end
    a = a(1:max_n);
end
function ok = consistency(a, n)
    v = a([1:n] + a(1:n));
    ok = (n == length(unique(v)));
end % Thomas Scheuerle, Nov 21 2023

Formula

From Thomas Scheuerle, Nov 26 2023: (Start)

Conjectures:

a(n) = A049472(n) = floor(n*(1 + 1/sqrt(2))) - n, if n is not in A083051.

a(A083051(n)) = A184119(n+1) - A083051(n).

a(a(A083051(n)) + A083051(n)) + a(A083051(n)) + A083051(n) = A328987(n) = floor((a(A083051(n)) + A083051(n))*(1 + 1/sqrt(2))) = floor(A184119(n+1)*(1 + 1/sqrt(2))). (End)