A random walk on a finite square with prime numbers

your conjecture is True assuming that : there is a $k$ sequence of consecutive primes that the difference between every two consecutive prime of that sequence produces the sequence $\{n-1,n-1,n-1,n-2,n-2,n-3,n-3,.....,2,2,1,1\} \mod n$ such that $n-1$ element is repeated $3$ time and all the other elements are repeated $2$ times in descending order.

which is $2n-1$ elements or $2n$ consecutive prime.

to explain what i mean, lets take $n=3$ so i am looking for $2*3= 6$ consecutive prime numbers that their consecutive difference produces $\{2,2,2,1,1\} \mod 3$

now if i assume that this will happen for all odd $n$ numbers, then no matter at what cell in the square i was standing or in what direction i was going, i am guaranteed to travel across every cell in the $n*n$ square.

note : as i wrote in the comments, i think that the assumption could be proved by Green-Tao theorem (i am not sure of that).

i hope this give you some idea on how to tackle this problem