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I was recently reading a post on this forum which said that there must be 15 captured pieces in order for there to be 6 pawns on the a-file or h-file. That got me thinking how complicated that would be, which got me thinking how many moves would it take for such a thing to happen. So, what is the minimum length of a legal game (i.e. reachable from starting position) where there are 6 pawns of the same color on either the a-file or h-file?

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This has already been covered over on Puzzling Stack Exchange. Courtesy of @xnor, the shortest possible game is in 31 moves. You can read the proof over there.

[FEN ""]

1. Nf3 Nf6 2. Nd4 Nd5 3. Nc6 dxc6 4. g4 Bf5 5. gxf5 h5 6. f4 Nd7 7. e4 Ne5 8. fxe5 Qd6 9. exd6 exd6 10. exd5 g6 11. Na3 Bg7 12. Rb1 Bd4 13. h4 Bc5 14. d4 g5 15. hxg5 Rh6 16. dxc5 dxc5 17. Bd2 Re6+ 18. fxe6 fxe6 19. Ba6 exd5 20. c4 dxc4 21. Ba5 Ke7 22. b4 cxb4 23. g6 bxa3 24. Rb6 bxa6 25. Rxh5 cxb6 26. Rb5 bxa5 27. Qb3 cxb3 28. g7 bxa2 29. g8=Q cxb5 30. Qg4 Ke8 31. Qa4 bxa4

On a similar tangent, on the retrograde analysis side, here is a fun little problem in which six pawns pile up on the e-file in a unique move sequence.

[Title "Nicolas Dupont, Problemesis 06/2002, Proofgame In 20.5 Moves"] 
[FEN ""]

1. h4 g5 2. hxg5 Nf6 3. gxf6 Bh6 4. fxe7 Be3 5. dxe3 b5 6. Qd6 b4 7. Qg3 b3 8. axb3 c5 9. Ra4 c4 10. bxc4 d5 11. cxd5 Be6 12. dxe6 Qd5 13. Rah4 Qf3 14. gxf3 Rg8 15. Bg2 Rg4 16. Kf1 Re4 17. fxe4 a5 18. f4 a4 19. Nf3 Ra5 20. Rg1 Re5 21. fxe5
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