A346268 a(n) is the smallest positive number not yet in a(0..n-1) such that the absolute differences of order k = 1 .. n-1 of a(0..n) contain a maximum of k-1 duplicate values. We define a(0) = 1.

1, 2, 4, 7, 12, 3, 9, 20, 40, 36, 5, 26, 18, 47, 115, 13, 6, 19, 44, 94, 193, 87, 60, 48, 84, 170, 355, 31, 14, 63, 25, 119, 71, 187, 444, 34, 8, 42, 98, 211, 450, 100, 10, 62, 132, 116, 274, 763, 50, 22, 73, 244, 792, 92, 1502, 5433, 27, 17, 41, 117, 77, 206, 540, 1315

Offset: 0

Author

Thomas Scheuerle, Jul 12 2021

Keywords

• nonn

Comments

Absolute differences of order 1 means k1 = {abs(a(0)-a(1)), abs(a(1)-a(2)), ...} and of order 2: k2 = {abs(k1(0)-k1(1)), abs(k1(1)-k1(2)), ...}.

If we did not allow any duplicated values in k1, k2, ..., kn we would get a case where all k1 .. kn would be the same sequence and this sequence would be s(n) = 2^n. In the case of this sequence k1..kn are not equal but there is still a remarkably strong correlation between a(n), k1(n) and kn(n).

It appears that if a(n) = p then the greatest possible number a(n-1) would be 1 + p + 2^p. If true this would have the consequence that this sequence would be a permutation of the positive integers, because in the case of a(n) = p, n could then not be greater than p + 2^p.

It appears that the logarithmic plot of this sequence consists of straight-line segments attached to each other; this indicates intervals of exponential growth.

If this sequence is a permutation of positive integers, will all k1 .. kn then contain all positive integers at least once?

Examples

			a(0..9) = {1,2,4,7,12,3,9,20,40,36} no duplicates.
k1(0..9) = {1,2,3,5,9,6,11,20,4,31} no duplicates.
k2(0..9) = {1,1,2,4,3,5,9,16,27,10} one duplicate 1.
k3(0..9) = {0,1,2,1,2,4,7,11,17,3} two duplicates 1 and 2.
		

Links

• Thomas Scheuerle, Log plot of a(0..200)(blue), k1(0..200)(yellow) and k2(0..200)(orange)

Crossrefs

Cf. A035313, A327743, A005282.

Programs

MATLAB

function a = A346268(max_n)
a(1) = 1;
t_min = 2;
    for n = 1:max_n
        t = t_min;
        while ~isok([a t])
            t = t+1;
        end
        a = [a t];
        if t == t_min+1
            t_min = t+1;
        end
    end
end
function [ ok ] = isok( num )
    ok = (length(num) == length(unique(num)));
    dnum = num;
    if ok
        for k = 1:(length(num)-1)
            dnum = abs(diff(dnum,1));
            ok = ok && ((length(dnum) - length(unique(dnum))) < k);
            if ~ok
                break;
            end
        end
    end
end