SB(3,n) has no Hamiltonian cycle for even n, proven by a sign-of-permutation obstruction that extends to SB(m,n) for all odd m congruent to 3 mod 4.
Knuth,The Art of Computer Programming, Volume 4, Pre-Fascicle 8a: Generating All Compositions and Subroutines for Combinatorial Searches(draft of 10 April 2026)
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$\mathit{SB}(3,n)$ has no Hamiltonian cycle when $n$ is even: a sign-of-permutation proof, with extension to all odd $m\equiv 3\pmod 4$
SB(3,n) has no Hamiltonian cycle for even n, proven by a sign-of-permutation obstruction that extends to SB(m,n) for all odd m congruent to 3 mod 4.