Distance formula for generalized knight movement on infinite chessboard from a corner Consider a chessboard infinite in positive x and y directions, all square has non-negative integer coordinates, and the only corner is at $(0,0)$. A $(p,q)$-knight is a piece that can move so that after each move one of the coordinate change by $p$ and the other change by $q$ (we will just call it a knight from now on). Set a knight at the corner $(0,0)$, and assume that $(p,q)$ is such that every position on the board can be reached by the knight.
For a position $(m,n)$ on the board, let $d(m,n)$ be the minimum number of moves needed for a knight from the corner to reach $(m,n)$.
Now the following claims are true:
$\gcd(p,q)=1$ and $p,q$ are not both odd. This is necessary and sufficient conditions for every square to be reachable. Necessary is easily seen, for sufficient a sketch of the solution is in this question Can an $(a,b)$-knight reach every point on a chessboard?
For every square on the board, every ways to reach it require the number to moves to have the same parity as $m+n$, this is from black-white coloring. So $d(m,n)$ has the same parity as $m+n$
$d(m,n)\max(p,q)>=\max(m,n)$, obviously.
$d(m,n)(p+q)>=m+n$
So let's $B(m,n)$ be the smallest integer that satisfy all the constraints: $B(m,n)\max(p,q)>=\max(m,n)$ and $B(m,n)(p+q)>=m+n$ and $B(m,n)$ has the same parity as $m+n$. Then we know that $d(m,n)>=B(m,n)$ for all $(m,n)$. We make $B(m,n)$ the predicted value of $d(m,n)$.
DEFINITION: An "awkward spot" on the board is a position $(m,n)$ in which $d(m,n)$ is not equal to $B(m,n)$.
QUESTION: is it true that for all valid values of $(p,q)$ then the number of awkward spots are finite?
Example: for the normal chess knight $(p,q)=(1,2)$ then you can check against this answer chess board knight distance (but need some small modification since we start from a corner) to see that the awkward spots are $(0,1),(1,0),(1,1),(2,2)$ so there are only a finite number of them.
(I have heard suggestions to use Fourier transform but I have no clues what to do with it)
 A: If I understand you correctly, the number of "awkward spots" can easily be infinite.  This is mainly because, in a sense, your definition of $B(m,n)$ is a bit too "optimistic".
Consider $(p,q) = (1,10)$.
Obviously any square $(k, 10k)$ can be reached in exactly $k$ moves.  What about $(k, 10k-2)$, for $k \ge 1$?  We have $B(k, 10k-2) = k$ because:


*

*$k \max(1,10) = 10k  \ge \max(k,10k-2) = 10k-2$

*$k (1 + 10) = 11k \ge k + (10k-2) = 11k - 2$

*$k$ has same parity as $k + (10k-2)$

*OTOH $(k-1) (1 + 10) = 11(k-1) < 11k -2$
However, the square $(k, 10k-2)$ cannot be reached in $k$ moves (or fewer for that matter), because:


*

*If all $k$ moves are of the form $(\pm 1, +10)$ then the final $y$-coordinate would be $10k$ and not $10k-2$.

*If at least one move is not $(\pm 1, +10)$ then the final $y$-coordinate is at most $10(k-1) + 1 = 10k -9 < 10k-2$.
Conclusion: For the $(1,10)$-knight, $(k, 10k-2)$ (and many similar squares) are awkward for any $k \ge 1$.

Further thoughts: in general for a $(p,q)$-knight to move to row $r$ (regardless of column) already requires something like solving the Bezout identity $px + qy = r$ with "minimum" coefficients $(x,y)$, in a sense.  My example shows that forgetting this bound already makes your $B(m,n)$ too optimistic.  A more interesting question is if you somehow include this in the definition of $B(m,n)$, then are there infinite number of awkward squares?  I don't know the answer to that question.
