For each n≥ 2, let G_n be the set of Gilbreath sequences of length n, let K_S denote the valid-extension set associated with S, and let U_n be the doubling sequence. Maximum-width conjecture. max_S∈G_n|K_S|=2^n-1+1, and the maximum is uniquely achieved by U_n. The doubling sequence attains the asserted value, but the source presents uniqueness and maximality as a conjecture; no resolution is supplied.

Maximum-width conjecture for valid-extension sets of Gilbreath sequences

Less than 1 year old · traced to 2026

For each $n\geq 2$ , let $\mathcal{G}_n$ be the set of Gilbreath sequences of length $n$ , let $K_S$ denote the valid-extension set associated with $S$ , and let $U_n$ be the doubling sequence. Maximum-width conjecture.

\max_{S\in\mathcal{G}_n}|K_S|=2^{n-1}+1,

and the maximum is uniquely achieved by $U_n$ . The doubling sequence attains the asserted value, but the source presents uniqueness and maximality as a conjecture; no resolution is supplied.

References

Primary source

Leila Muney, “Holes in Valid-Extension Sets of Finite Gilbreath Sequences”, arXiv:2606.23721 (2026).

combinatorics integer-sequences

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A 2026 paper proves the doubling sequence reaches the proposed size and checks the claim through length ten, while a posted complete proof remains unverified.

The conjecture asks whether, for every $n\geq 2$ , the largest valid-extension set has size $2^{n-1}+1$ and whether the doubling sequence $U_n$ is the unique maximizer. Leila Muney’s June 2026 preprint records this as Conjecture 30, not as a theorem.

Known results

Muney, 2026: $|K_{U_n}|=2^{n-1}+1$ for every $n\geq 2$ .
Muney, 2026: exhaustive computation verifies maximality and uniqueness for $2\leq n\leq 10$ .
The preprint gives no general proof, counterexample, or resolution.

Posted attempt

A reader claims a complete proof: a universal anti-diagonal bound yields $|K_S|\leq 2^{n-1}+1$ , and equality forces the doubling sequence. The argument has not been independently verified, so it does not establish the conjecture.

Current status (as of August 2026): The doubling sequence’s value and the cases $n\leq 10$ are established, but general maximality and uniqueness remain unverified despite a posted complete-proof claim.

Sources

arxiv.org

ProofThis solution needs a summarySee full solutionHide full solution

Let

S=(s_1,\ldots,s_n)=(2,3,s_3,\ldots,s_n)

be a strictly increasing Gilbreath sequence, and form its difference triangle

s_j^0=s_j,\qquad s_j^{b+1}=|s_{j+1}^b-s_j^b|.

The defining Gilbreath condition is

s_1^b=1\qquad(1\le b<n).

We first establish the universal column bound

\boxed{\quad s_j^b\le2^{j-1} \qquad(b\ge1,\ j+b\le n). \quad} \tag{1}

For $j=1$ , this is the defining condition. Suppose the bound holds throughout column $j$ . Whenever $s_{j+1}^b$ exists, the triangle recurrence gives

s_{j+1}^b \le s_j^b+s_j^{b+1} \le2^{j-1}+2^{j-1} =2^j.

Thus (1) follows by induction.

Let

e_i=s_{n-i}^i,\qquad A(S)=\sum_{i=1}^{n-1}e_i.

Applying (1) to the right anti-diagonal yields

e_i\le2^{n-i-1},

and consequently

A(S) \le\sum_{i=1}^{n-1}2^{n-i-1} =2^{n-1}-1. \tag{2}

The valid-extension set $K_S$ is contained in the candidate set $C_S$ , and Lemma 7 gives

|C_S|=A(S)+2.

Therefore

|K_S| \le |C_S| =A(S)+2 \le2^{n-1}+1. \tag{3}

Theorem 29 shows that the doubling sequence

U_n=(2,3,5,9,17,\ldots,2^{n-1}+1)

attains this bound. Hence

\max_{S\in\mathcal G_n}|K_S|=2^{n-1}+1.

It remains to prove uniqueness. If equality holds in (3), equality holds in (2), and each individual anti-diagonal bound must be attained. In particular,

s_{n-1}^1=e_1=2^{n-2}.

Whenever $s_j^1=2^{j-1}$ , the recurrence and (1) give

2^{j-1} =s_j^1 \le s_{j-1}^1+s_{j-1}^2 \le2^{j-2}+2^{j-2} =2^{j-1}.

Equality throughout forces

s_{j-1}^1=2^{j-2}.

Descending induction therefore gives

s_j^1=2^{j-1}\qquad(1\le j<n).

Because $S$ is strictly increasing, its first difference row consists of its actual positive gaps:

s_{j+1}-s_j=s_j^1=2^{j-1}.

Starting with $s_1=2$ , we obtain

s_j=2^{j-1}+1\qquad(2\le j\le n).

Thus $S=U_n$ , proving that the maximizer is unique for every $n\ge2$ .