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(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?
LINKS
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
KEYWORD
nonn
AUTHOR
Thomas Scheuerle, Jul 12 2021
STATUS
approved