Improve upper bound for Davenport constant C_{53} to 4

[maaxgrin]

This PR updates the entry for the Davenport constant of $C_n^3$, improving the recorded upper bound

$$ C_{53}\le 20369 $$

to

$$ C_{53}\le 4. $$

The update records the stronger pointwise estimate

$$ D(C_n^3)\le 4n-P(n)-2, \qquad P(n)=\max_{p^a\parallel n}p^a, $$

which implies $D(C_n^3)\le 4(n-1)$ for all $n\ge 2$.

Proof details

Let $D_k(G)$ denote the $k$-wise Davenport constant. We use the following consequence of the estimates of Freeze--Schmid and Bhowmik--Schlage-Puchta.

Local lemma. For every prime $p$ and every $k\ge 3p-2$,

$$ D_k(C_p^3)\le pk+p^2. $$

For $p=2$, Freeze--Schmid give $D_k(C_2^3)=2k+3$ for $k\ge2$, so $D_k(C_2^3)\le2k+4$.

For $p=3$, Bhowmik--Schlage-Puchta use $\eta(C_3^3)=17$ and the fact that length $15$ already forces three disjoint zero-sums, hence $D_3(C_3^3)\le17$. Repeatedly extracting zero-sums of length at most $3$ gives

$$ D_k(C_3^3)\le17+3(k-3)=3k+8\le3k+9. $$

For $p=5$, Bhowmik--Schlage-Puchta give $\eta(C_5^3)=33$ and $D_3(C_5^3)\le33$, hence

$$ D_k(C_5^3)\le33+5(k-3)=5k+18\le5k+25. $$

For $p\ge7$, we use the $d=3$ part of Bhowmik--Schlage-Puchta, Lemma 2.3 and the proof of Theorem 1.4. Set

$$ M=(p+1)(3p-7)+4=3p^2-4p-3. $$

They give $\eta(C_p^3)\le M$ and the multi-wise estimate

$$ D_j(C_p^3)\le \max{5p-2,\frac{3(p-1)}2j+2p+5}. $$

Take

$$ j_0=\left\lfloor \frac{M-2p-5}{\frac32(p-1)} \right\rfloor. $$

Then $D_{j_0}(C_p^3)\le M$, $j_0<2p<3p-2$, and $M\le p^2+pj_0$. Since $M\ge\eta(C_p^3)$, repeated extraction of short zero-sums gives, for every $k\ge3p-2$,

$$ D_k(C_p^3)\le M+p(k-j_0)\le pk+p^2. $$

This proves the local lemma.

The core of the proof is the following induction, which starts from the largest primary component of $n$ instead of starting from $1$.

Let $Q=P(n)$, and write

$$ n=Qp_1\cdots p_s, $$

where the remaining prime factors are counted with multiplicity. Since $Q$ is the largest primary component of $n$, every $p_i\le Q$. Define

$$ m_0=Q,\qquad m_i=Qp_1\cdots p_i, $$

For these $m_i$, we prove by induction that

$$ D(C_{m_i}^3)\le 4m_i-Q-2. $$

For $i=0$, $m_0=Q$ is a prime power, so the $p$-group formula gives

$$ D(C_Q^3)=3Q-2=4Q-Q-2. $$

For the induction step, put $m=m_{i-1}$ and $p=p_i$. In $C_{pm}^3$, take the natural subgroup $H\simeq C_m^3$, so $C_{pm}^3/H\simeq C_p^3$. The standard inductive inequality gives

$$ D(C_{pm}^3)\le D_{D(C_m^3)}(C_p^3). $$

Also $D(C_m^3)\ge 3m-2\ge 3p-2$, since $m\ge Q\ge p$. Thus the local estimate applies with $k=D(C_m^3)$, giving

$$ D(C_{pm}^3)\le pD(C_m^3)+p^2. $$

Using the induction hypothesis,

$$ D(C_{pm}^3) \le p(4m-Q-2)+p^2. $$

Since $p\le Q$, this is at most $4pm-Q-2$, because the required inequality is

$$ (p-1)Q\ge p^2-2p+2. $$

This follows from $(p-1)Q\ge p(p-1)\ge p^2-2p+2$.

This proves the induction. Taking $m_s=n$, we get

$$ D(C_n^3)\le 4n-Q-2=4n-P(n)-2. $$

Since $P(n)\ge2$, this implies $D(C_n^3)\le4(n-1)$, and hence $C_{53}\le4$.

Files changed:

• README.md: updates the table upper bound for $C_{53}$ from 20369 to 4.
• constants/53a.md: adds the pointwise bound, upper-bound table entry, and references.

AI assistance disclosure: the proof of the new upper bound was found by GPT-5.5 Pro; the references and proof details were checked against the cited sources before submission.

[maaxgrin]

I added a conditional Lean 4 formalization of the proof:

RequestProject/Main.lean

It compiles with Lean 4.28.0 / mathlib v4.28.0. The file isolates the results from cited previous works as axioms and checks the local estimate, the induction step, and the final bound.

The formalization was prepared with the help of Aristotle and GPT-5.5 Pro.