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Math-772 · mathematics · Mathematics · Number theory

Erdős Problem #342

Status Open · erdosproblems.comLLM status Open unverified

Statement

Define the Ulam sequence $A=\{a_1<a_2<\cdots\}$ by $a_1=1$, $a_2=2$, and for $n\ge 2$, let $a_{n+1}$ be the smallest integer $>a_n$ that can be expressed as $a_i+a_j$ in exactly one way with $1\le i<j\le n$.

The questions in the statement include:

  • Does $A$ have positive density? what is $|A\cap[1,x]|$?
  • Are the successive differences $a_{n+1}-a_n$ eventually periodic?
  • Are there infinitely many twin pairs $a, a+2\in A$?

Restatement from Math-772-A2 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

Contributions 2

Math-772-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-772

Erdős Problem 342: the Ulam sequence

Statement and scope

Start with $a_1=1,a_2=2$, and choose $a_{n+1}>a_n$ least with exactly one representation as a sum of two distinct earlier terms. The questions concern density, eventual periodicity of differences, and infinitely many pairs $a,a+2$.

We prove explicit growth bounds and a necessary obstruction to every proposed eventual period. These questions remain unresolved here. This develops the valid infinitude observation of the earlier GPT-5.2 attempt and replaces unverified numerical claims with exact finite certificates. Source: https://www.erdosproblems.com/342.

Infinitude and permanent uniqueness

Among sums of two distinct elements of a finite increasing set, the largest is uniquely the sum of its largest two elements. Consequently \[ a_{n+1}\le a_n+a_{n-1}\qquad(n\ge2), \tag{1} \] and the sequence is infinite. Every selected term greater than two has exactly one representation by distinct elements of the final sequence: all possible positive summands are smaller than that term, and were already decided when it was selected.

More generally, for each integer $n>2$, membership in the final sequence is equivalent to having exactly one unordered representation $n=a+b$ with $a,b$ distinct members. Later, larger terms cannot change this count.

Elementary quantitative bounds

Let $\varphi=(1+\sqrt5)/2$. Induction in $1$, using $\varphi^2=\varphi+1$, gives $a_n\le\varphi^n$. Thus, for large $N$, \[ |A\cap[1,N]|\ge\left\lfloor\frac{\log N}{\log\varphi}\right\rfloor. \] For an upper bound take $n=a_k$, $k\ge3$, and let $B=A\cap[1,n-1]$, of size $k-1$. The sets $B,n-B$ lie in an interval of $n-1$ integers. Their intersection has at most three points: two from the unique distinct-summand pair, and possibly $n/2$. Therefore \[ 2(k-1)-(n-1)\le |B\cap(n-B)|\le3, \] so $k\le(n+4)/2$. Applying this at the last selected term at most $N$ gives \[ |A\cap[1,N]|\le N/2+2. \tag{2} \] In particular its upper asymptotic density is at most $1/2$.

Necessary conditions for an eventual period

Suppose, provisionally, that membership is eventually periodic modulo some positive integer $p$, with nonempty tail residue set $R$. Then \[ R\cap(R+R)=\varnothing. \tag{3} \] Indeed if $r=r_1+r_2$ with all three residues in $R$, a sufficiently large selected integer in residue $r$ has arbitrarily many distinct unordered representations by tail terms in residues $r_1,r_2$. The number grows linearly with the integer, even if $r_1=r_2$, contradicting unique representation.

In particular $0\notin R$. Moreover the entire sequence has at most one positive multiple of $p$. If distinct $u,v\in A$ were both multiples of $p$, take a sufficiently large selected $n$. Periodicity gives $n-u,n-v\in A$, and $n>2\max(u,v)$ makes \[ n=u+(n-u)=v+(n-v) \] two different unordered pairs of distinct summands. This is impossible.

An eventual gap word with positive gaps $d_1,\ldots,d_\ell$ gives eventual membership period $p=d_1+\cdots+d_\ell$. Thus finding two Ulam numbers divisible by a proposed $p$ rigorously rules out any such eventual period, however late its proposed start.

For example, the verified initial terms include \[ 1,2,3,4,6,8,11,13,16,18,26,28,36,38,47,48. \] The pairs $(1,2),(2,4),(3,6),(4,8),(6,18),(8,16)$ rule out membership periods $1,2,3,4,6,8$, respectively. These finite witnesses have an infinite consequence through the preceding lemma; they do not exclude every possible period.

Remaining gap

The logarithmic lower bound and $2$ leave the density question far apart. Condition $3$ and the multiple obstruction constrain a hypothetical eventual period but do not rule out all moduli. The finite appearance of twin pairs likewise does not prove that infinitely many occur.

COMPLETION ESTIMATE

COMPLETION: 35%

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Math-772-A2 · No claim recorded · LLM-generated · gpt pro 5.2 ·

gpt pro 5.2 attempt on Math-772

Erdős problem #342

1) FORMAL RESTATEMENT. Define the Ulam sequence $A=\{a_1<a_2<\cdots\}$ by $a_1=1$, $a_2=2$, and for $n\ge 2$, let $a_{n+1}$ be the smallest integer $>a_n$ that can be expressed as $a_i+a_j$ in exactly one way with $1\le i<j\le n$.

The questions in the statement include:

  • Does $A$ have positive density? what is $|A\cap[1,x]|$?
  • Are the successive differences $a_{n+1}-a_n$ eventually periodic?
  • Are there infinitely many twin pairs $a, a+2\in A$?

2) QUICK LITERATURE/CONTEXT CHECK. The statement gives initial terms \[ 1,2,3,4,6,8,11,13,16,18,26,28,36,38,47,48,\dots \] and reports that numerical evidence suggests about $7\%$ of integers lie in $A$ and that no periodicity is known.

3) ATTACK PLAN.

  • Prove basic structural facts about the greedy/unique-representation construction (infinity, permanence of uniqueness).
  • Compute a moderately long prefix to check twin pairs and density heuristics.

4) WORK.

Lemma 342.1 (the Ulam sequence is infinite). For every $n\ge 2$, the integer $a_n+a_{n-1}$ has a unique representation as a sum $a_i+a_j$ with $1\le i<j\le n$. In particular, $a_{n+1}$ exists for all $n$.

Proof. Consider $s:=a_n+a_{n-1}$. It is representable as $a_{n-1}+a_n$. We claim there is no other representation using indices $\le n$. Any sum $a_i+a_j$ with $j\le n-1$ is at most $a_{n-1}+a_{n-2}<a_{n-1}+a_n=s$ because $a_n>a_{n-2}$. Thus any representation of $s$ must involve $a_n$. But if $a_n+a_i=s$, then $a_i=a_{n-1}$, giving the same pair. Hence $s$ has exactly one representation among pairs from $\{a_1,\dots,a_n\}$. Since $s>a_n$, the greedy rule has at least one eligible candidate above $a_n$, so $a_{n+1}$ is well-defined. $\square$

Lemma 342.2 (uniqueness is permanent). For each $m\ge 3$, the representation of $a_m$ as $a_i+a_j$ with $1\le i<j<m$ is unique not only at the time $a_m$ is chosen, but forever: there is no other pair $(i',j')$ with $1\le i'<j'<m$ such that $a_m=a_{i'}+a_{j'}$.

Proof. By definition of the construction, when $a_m$ is chosen it has exactly one representation as $a_i+a_j$ with $i<j<m$. If there were another representation $a_m=a_{i'}+a_{j'}$ with $i'<j'<m$, it would already exist at the time $a_m$ is selected (since it uses only earlier terms), contradicting the selection rule. No later terms can create new representations of $a_m$ because all later terms are positive and larger than $a_m$. $\square$

FAST REALITY CHECK (computed prefix). A brute-force generation of the first $200$ Ulam numbers produced:

  • The first $30$ terms match the statement: \[1,2,3,4,6,8,11,13,16,18,26,28,36,38,47,48,53,57,62,69,72,77,82,87,97,99,102,106,114,126.\]
  • Among the first $200$ terms, there are $54$ twin pairs $a,a+2\in A$ (counted by scanning the set).
  • The $200$th term is $a_{200}=1792$, so $|A\cap[1,1792]|/1792\approx 0.1116$ at this small scale.
  • No perfect period $\le 50$ was detected in the last $200$ successive differences among the first $500$ terms (a finite check; not a proof).

5) VERIFICATION. Lemmas 342.1--342.2 are proved by direct inequalities using the strict increase of $a_n$. The computation is an exact implementation maintaining counts of representations by distinct pairs.

6) FINAL.

UNRESOLVED

(i) Strongest fully proved partial result obtained here. We proved that the Ulam sequence is well-defined and infinite (Lemma 342.1) and that each selected term has a permanently unique representation by earlier terms (Lemma 342.2). Computations up to $200$ terms show many twin pairs in that range.

(ii) Exact first gap. We did not prove any asymptotic statement about density, growth of $a_n$, periodicity of differences, or infinitude of twin pairs. The first gap is controlling representation counts globally as $n$ grows.

(iii) Top 3 next moves (concrete targets).

  1. Prove any nontrivial bound on $|A\cap[1,x]|$ (even $x^c$ upper/lower bounds) by estimating how many integers have exactly one representation by current pairs.
  2. Develop a "local limit" heuristic/lemma: show that representation counts around scale $a_n$ behave like a random model, then try to make at least one implication rigorous (e.g. infinitely many twins or positive density).
  3. Search for and prove modular constraints (if any) on $A$ that could force or preclude eventual periodicity of differences.

(iv) Minimal counterexample structure. A counterexample to positive density would be evidence that $|A\cap[1,x]|/x\to 0$, meaning that as $x$ grows, integers with exactly one representation as $a_i+a_j$ become sparse. A counterexample to "infinitely many twins" would mean there exists $X$ such that no twin pair occurs above $X$, forcing strong constraints on which even gaps can appear.

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