Monotonicity conjecture for normalized mixed-code sphere sizes
Let be the alphabet length, let denote the size of the sphere of radius , and let be the binomial coefficient. The quantities are defined for . Monotonicity conjecture. The sequence
is decreasing for . This would extend the endpoint bounds, which are attained at and , and give a stronger monotonicity principle for the sphere sizes of mixed codes with finite alphabets.
References
Primary source
Yonatan Yehezkeally, Haider Al Kim, Sven Puchinger and Antonia Wachter-Zeh, “Bounds on Mixed Codes with Finite Alphabets”, arXiv:2212.09314 (2022).
Progress summary
A reader-written complete proof claims to settle the conjecture for every alphabet choice, but the claim has not been independently verified.
Yonatan Yehezkeally, Haider Al Kim, Sven Puchinger, and Antonia Wachter-Zeh stated the conjecture in 2022. It asks whether the normalized sphere sizes decrease with ; their paper explicitly labels this Conjecture 5.
Known results
- Theorem 3 proves that is nonincreasing in .
- Theorem 4 proves endpoint bounds for , with equality at and .
- The paper gives no proof or counterexample for the conjectured root monotonicity.
Posted attempt
A reader-written argument claims a complete proof for all finite alphabet sizes, deriving the desired inequalities from Theorem 3. This attempt has not been independently verified.
Current status (as of August 2026): The conjecture has a complete-proof claim but no independent verification; the endpoint bounds and related ratio monotonicity are settled, while the conjecture itself remains unconfirmed.
Sources
Solutions 1
ProofThis solution needs a summarySee full solution
Complete proof. Let , put , and let be the mixed-code sphere size. Then
Define
Theorem 3 of the source proves that
is nonincreasing as a function of .
For , it follows that
because every factor with is at least . Therefore
and hence
This is exactly the conjectured monotonicity for all alphabet sizes. Equivalently, it is Maclaurin's inequality for the positive numbers . Equality occurs precisely in the homogeneous case .