Monotonicity conjecture for normalized mixed-code sphere sizes

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Let nn be the alphabet length, let srs_r denote the size of the sphere of radius rr, and let (nr)\binom{n}{r} be the binomial coefficient. The quantities sr(nr)\frac{s_r}{\binom{n}{r}} are defined for 1⩽r⩽n1\leqslant r\leqslant n. Monotonicity conjecture. The sequence

(sr(nr))1/r\left(\frac{s_r}{\binom{n}{r}}\right)^{1/r}

is decreasing for 1⩽r⩽n1\leqslant r\leqslant n. This would extend the endpoint bounds, which are attained at r=1r=1 and r=nr=n, 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

Refreshed
Claimed solved

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 (sr/(nr))1/r\left(s_r/\binom{n}{r}\right)^{1/r} decrease with rr; their paper explicitly labels this Conjecture 5.

Known results

  • Theorem 3 proves that (r+1)sr+1(n−r)sr\frac{(r+1)s_{r+1}}{(n-r)s_r} is nonincreasing in rr.
  • Theorem 4 proves endpoint bounds for sr/(nr)s_r/\binom{n}{r}, with equality at r=1r=1 and r=nr=n.
  • 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 solutionHide full solution

Complete proof. Let q1,…,qn≥2q_1,\ldots,q_n\ge2, put ai=qi−1>0a_i=q_i-1>0, and let srs_r be the mixed-code sphere size. Then

sr=∑I⊆{1,…,n}∣I∣=r∏i∈Iai=er(a1,…,an).s_r=\sum_{\substack{I\subseteq\{1,\ldots,n\}\\|I|=r}} \prod_{i\in I}a_i=e_r(a_1,\ldots,a_n).

Define

Er=sr(nr)(0≤r≤n),E0=1.E_r=\frac{s_r}{\binom nr}\quad(0\le r\le n), \qquad E_0=1.

Theorem 3 of the source proves that

dr=(r+1)sr+1(n−r)sr=Er+1Er,0≤r<n,d_r=\frac{(r+1)s_{r+1}}{(n-r)s_r} =\frac{E_{r+1}}{E_r}, \qquad 0\le r<n,

is nonincreasing as a function of rr.

For 1≤r<n1\le r<n, it follows that

Er=∏j=0r−1dj≥dr r,E_r=\prod_{j=0}^{r-1}d_j\ge d_r^{\,r},

because every factor djd_j with j<rj<r is at least drd_r. Therefore

Er+1=Erdr≤Er 1+1/r,E_{r+1}=E_rd_r\le E_r^{\,1+1/r},

and hence

(sr+1(nr+1))1/(r+1)≤(sr(nr))1/r(1≤r<n).\boxed{\left(\frac{s_{r+1}}{\binom n{r+1}}\right)^{1/(r+1)} \le \left(\frac{s_r}{\binom nr}\right)^{1/r}} \qquad(1\le r<n).

This is exactly the conjectured monotonicity for all alphabet sizes. Equivalently, it is Maclaurin's inequality for the positive numbers qi−1q_i-1. Equality occurs precisely in the homogeneous case q1=⋯=qnq_1=\cdots=q_n.