

A346268


a(n) is the smallest positive number not yet in a(0..n1) such that the absolute differences of order k = 1 .. n1 of a(0..n) contain a maximum of k1 duplicate values. We define a(0) = 1.


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
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OFFSET

0,2


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(n1) 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 straightline 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?


LINKS

Table of n, a(n) for n=0..63.
Thomas Scheuerle, Log plot of a(0..200)(blue), k1(0..200)(yellow) and k2(0..200)(orange)


EXAMPLE

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.


PROG

(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


CROSSREFS

Cf. A035313, A327743, A005282.
Sequence in context: A153555 A259588 A058103 * A097592 A267699 A193841
Adjacent sequences: A346265 A346266 A346267 * A346269 A346270 A346271


KEYWORD

nonn


AUTHOR

Thomas Scheuerle, Jul 12 2021


STATUS

approved



