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

From papers

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.

Progress summary

Open

A June 2026 paper proves the proposed value for the doubling sequence and checks the conjecture through length ten, but does not settle the general case.

The conjecture says that, for every $n\geq 2$ , the largest valid-extension set has size $2^{n-1}+1$ and is uniquely attained by the doubling sequence $U_n$ . No proposer or original date is identified in the retrieved sources.

June 2026 preprint

The preprint proves $|K_{U_n}|=2^{n-1}+1$ for every $n\geq 2$ and reports exhaustive verification of maximality and uniqueness for $n\leq 10$ . It records the all- $n$ statement as Conjecture 30, noting that a proof must exploit the fact that the relevant anti-diagonal comes from a Gilbreath sequence; no proof, counterexample, or claimed resolution was found.

Current status (as of August 2026): The doubling sequence is proved to attain $2^{n-1}+1$ , and the conjecture is verified for $n\leq 10$ ; maximality and uniqueness for general $n$ remain open.

Sources

arxiv.org

Sources & referencesView supporting material

Primary source

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

combinatorics integer-sequences

1 week ago

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 $n\ge2$ .

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