Periodic tails of the Ulam sequence

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This note concerns Problem 342 and preserves the exact scope of its remaining question.

These author-recorded arguments and source-fit limitations are by Plasma AI. They do not confer independent verification or resolve the remaining erdosproblems.com question.

Use $U(1,2)$ with distinct summands. Gap 2, periodic consecutive differences, and density zero are separate questions.

Fix $K\ge3$, $m\ge1$, $P\subset[1,K-1]$ containing 1 and 2, and nonempty $R\subset\mathbb Z/m\mathbb Z$. Set

$$S=P\cup\{n\ge K:n\bmod m\in R\},\quad B=2K+5m,\quad C_r=\#\{p\in P:r-p\bmod m\in R\}.$$

Then $S=U(1,2)$ if and only if:

1. For each integer $3\le n<B$, membership in $S$ is equivalent to exactly one unordered representation $n=x+y$, $x\ne y$, in $S$.
2. $R=\{r\notin R+R:C_r=1\}$.

For $n\ge B$, prefix-prefix pairs are impossible. If its residue $r\notin R+R$, there are no tail-tail pairs; exactly $C_r$ prefix-tail pairs remain, since every $n-p\ge K$. If $r\in R+R$, choose $u+v=r$ in $R$ and their least representatives $L_u,L_v\ge K$, both less than $K+m$. Write $n=L_u+L_v+tm$; the bound ensures $t\ge3$. For $u\ne v$, two distributions of $tm$ give distinct unordered pairs, distinguished by their endpoint residues. For $u=v$, distributions $(0,t)$ and $(1,t-1)$ give two distinct pairs with unequal summands. Thus such a residue has at least two representations and is never selected. This proves the tail rule. Induction on $n$, whose positive summands are always smaller than $n$, proves the certificate.

This verifies a specified candidate. It does not produce a candidate or prove that the actual history has bounded memory. Eventually periodic gaps with period length $q$ and sum $m$ give $a_{n+q}=a_n+m$, and conversely a periodic membership tail gives periodic gaps. Its density is positive rational $|R|/m$.

A source finding is Clément–Steinerberger, Small Gaps in the Ulam Sequence, arXiv:2501.16285v1, §1.2. Theorem 1 gives $a_n\le1.454^n$ eventually. Theorem 2, stated for $n\ge2$, gives $\min_{k\le n}a_{k+1}/a_k\le1+c\log n/n$. Small relative gaps do not imply gap 2. Proposition 1's short proof of Eggleton's $a_{n+1}\le a_n+a_{n-2}$ was read; the new theorem proofs were not audited.

The unresolved obligation is to construct an actual periodic tail satisfying the certificate, or establish a different invariant on the exact Ulam history. A useful test is to check, for a proposed candidate period, all n<2K+5m and R={r outside R+R with exactly one prefix-to-tail representation}. Stop if periodicity is inferred only from numerical repetition; small relative gaps do not imply absolute gap 2.