An entire analysis of standard English Peg Solitaire has been given. See

Berlekamp, E. R.; Conway, J. H.; Guy, R. K. (2001) [1981], Winning Ways for your Mathematical Plays (paperback) (2nd ed.), A K Peters/CRC Press, ISBN 978-1568811307.

Consider the standard English version of the game. The center peg is adjacent to the center-most peg of the left-most column of a $ 3 \times 3$ grid of pegs on the right, the center-most peg of the right-most column of a $ 3 \times 3$ grid of pegs on the left, and similar adjacency in the up and down directions from the center peg. I’m curious whether a variation of the game has been considered with, say, $ 2n+1 \times 2n+1$ grids in the four directions from the center peg? Many questions could be asked that were thoroughly answered for the standard game where $ n=1$ , like can jumps be made so that there is only one peg remaining? If so, how many ways are there to win both with and without symmetry? Obviously there are $ 4(2n+1)^2 -4 +1=16n^2+16n+1$ pegs in this variation.

Thanks for your time!