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The record for the most mutual checks in a legal position is known. But what is the record for double checks?

There is one rule: threefold repetition applies. Four categories exist; promoted pieces allowed and not: and forced, i.e. Black defends, and unforced, i.e. Black helps.

Addendum: @Hauke Reddmann commented a known double checking mechanism. The origin of it is this century old problem as @Evargalo proposed a comment.

[Title "Alain Campbell White, Pittsburgh Gazette Times 4/1916, Mate In 12"]
[FEN "2q5/2pp4/3pr3/4pb2/K1p2pn1/2bn1kp1/3pr1R1/6BB w - - 0 1"]

1. Rf2+! Ke3 2. Rf3+ Ke4 3. Re3+ Kd4 4. Re4+ Kd55. Rd4+ Kc5 6. Rd5+ Kc6 7. Rc5+ Kb6 8. Rc6+ Kb7 9. Rb6+ Ka7 10. Rb7+ Ka8 11. Ra7+ Kb8 12. Ra8#

The record for the longest sequence in mating problem is 13, but it cannot contend for the record since Black’s king may alk out of the sequence.

[Title "Dragan Stojnic & Milomir Babic, The Problemist 2004-05, Mate In 13"]
[FEN "3q1nKB/R1P1kPRB/N3p3/1p1n2p1/2r2p2/1p2b3/P2pb2N/3r4 w - - 0 1"]

1. c8=N+! Kf6 2. Rg6+ Kf5 3. Rf6+ Ke5 4. Rf5+ Ke45. Re5+ Kd4 6. Re4+ Kd3 7. Rd4+ Kc3 8. Rd3+ Kc2 9. Rc3+ Kb2 10. Rc2+ Kb1 11. Rb2+ Ka1 12. Rb1+ Kxa2 13. Nb4#
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    Ba1 Bb1 Rb2 - Kc3 is a well known mechanism. 11 double checks. Also, the downvoters may have a point insofar as this SE is more dedicated to OTB than problem chess (still, I earned tons of undeserved upvotes by just quoting a position I happened to know from another forum, answering a problem chess question :-). That said, why don't you carry your question to the MatPlus forum, where the problem chess enthusiasts sit? They will be most interested (and also beat your records in no time :-)
    – Hauke Reddmann
    Commented Apr 8, 2019 at 7:54
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    Depending of how you handle threefold repetitions, the answer might be infinity, with W:Ka8,Ba1,Bh1,Re4 / B: Kd5 : 1.Rd4 Ke5 2.Re4 Kd5 3.Rd4 etc.
    – Evargalo
    Commented Apr 8, 2019 at 9:20
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    A (nice) problem from 2004 cannot be "the orginal problem in which [the double staircase] appears." I won't be surprised if you were off by a century !
    – Evargalo
    Commented Jun 5, 2019 at 12:42

2 Answers 2

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I am pretty sure I could make this last longer, but here is already a sequence with 25 unforced consecutive double-checks without promoted pieces.

[FEN "8/8/8/Rp2p3/3kP1N1/2N3pB/1Q5B/K3R3 w - - 0 1"]

1. Ne2+ Kxe4 2. Nc3+ Kd4 3. Ne2+ Ke4 4. Nc3+ Kd4 5. Nxb5+ Kd5 6. Nc3+ Kd4 7. Nb5+ Kd5 8. Nc3+ Kd4 9. Ne2+ Ke4 10. Nxg3+ Kf4 11. Ne2+ Ke4 12. Nc3+ Kd4 13. Nb5+ Kd5 14. Nc3+ Kd4 15. Ne2+ Ke4 16. Ng3+ Kf4 17. Ne2+ Kf5 18. Nh6+ Ke4 19. Ng3+ Kf4 20. Ne2+ Ke4 21. Nc3+ Kd4 22. Nb5+ Kd5 23. Nc3+ Kd4 24. Nb5+ Kd5 25.Nc7+

12 forced double-checks, based on the scheme given by Hauke Reddmann in a comment:

[FEN "1RN5/5p2/7p/3P3p/2P3p1/QPk4N/1R1p1K2/BB6 w - - 0 1"]

1. Rc2 Kd3 2. Rc3 Kd4 3. Rd3 Ke4 4. Rd4 Ke5 5. Re4 Kf5 6. Re5 Kf6 7. Rf5 Kg6 8. Rf6 Kg7 9. Rg6 Kh7 10. Rg7 Kh8 11. Rh7 {11. Rg8} Kg8 12.Ne7
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With promoted pieces, 18 forced double checks, in a matrix borrowed from the appendded selfmate, is possible.

[FEN "BQN3N1/1RPRPk2/1B6/1p4pp/2pp3n/4p3/4K1p1/1R2N3 w - - 0 1"]

1. e8=Q+ Kxe8 2. Nd6+ Kxd7 3. c8=B+ Kc6 4. Rc7+ Kxd6 5. Rc6+ Kd5 6. Rd6+ Ke5 7. Rd5+ Ke4 8. Re5+ Kf4 9. Rf5+ Kg4 10. Rf4+ Kg3 11. Rg4+ Kh3 12. Rg3+ Kh2 13. Rxg2+ Kh1 14. Rg1+ Kxg1 15. Nf3+ Kg2 16. Ne1+ Kg1 17. Nf3+ Kg2 18. Ne1+

The selfmate problem.

[Title "Miodrag Mladenovic, MatPlus.net Forum 14/11/2019, s#15"]
[FEN "BQN3N1/1RPRPk2/1B6/1b4np/2nr3r/4q3/4P1p1/4K3 w - - 0 1"]

1. e8=Q+ Kxe8 2. Nd6+ Kxd7 3. c8=B+ Kc6 4. Rc7+ Kxd6 5. Rc6+ Kd5 6. Rd6+ Ke5 7. Rd5+ Ke4 8. Re5+ Kf4 9. Rf5+ Kg4 10. Rf4+ Kg3 11. Rg4+ Kh3 12. Rg3+ Kh2 13. Rxg2+ Kh1 14. Rg1+ Kxg1 15. Qg3+ Qxg3#
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