The interesting case seems to be the even case. It seems like the lower half of the Alexander polynomial of the pretzel knot $ P(2m+1,2n,2k+1)$ , up to multiplication by $\pm t^{\alpha}$ , is given by   $$  \Delta_{h}(t)= -nt + \sum_{i=2}^{2m+1}(-1)^{i}( i+2n-1) t^{i} \\ +(2m+2n+1) \sum_{i=2m+2}^{k+m+2} (-1)^{i+1}  t^{i} $$ 

for $k\ge m\geq 0,n\ge 1.$ The subindex $h$ in $\Delta_{h}(t)$ is to indicate that it is half of the Alexander polynomial. We can assume  $k\ge m$ without loss of generality because we can exchange the parameters $2m+1$ and $2k+1$.