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Math-5679

Erdős Problem #481

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Statement

Let $a_i,b_i\in\mathbb N$ for $1\le i\le r$, with $C:=\sum_i1/a_i>1$. Starting from the one-element sequence $A_1=(1)$, replace each entry $x$ at every step by all $a_ix+b_i$, retaining multiplicities. We prove that some level $A_k$ has repeated entries. In particular, collisions at different levels alone would not suffice.

Restatement from Math-5679-A1 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

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Math-5679-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-5679

Erdős Problem #481: collisions under expanding affine maps

Complete harmonic-weight reconstruction, 24 September 2026.

Statement

Let $a_i,b_i\in\mathbb N$ for $1\le i\le r$, with $C:=\sum_i1/a_i>1$. Starting from the one-element sequence $A_1=(1)$, replace each entry $x$ at every step by all $a_ix+b_i$, retaining multiplicities. We prove that some level $A_k$ has repeated entries. In particular, collisions at different levels alone would not suffice.

Harmonic mass cannot grow both linearly and exponentially

Assume every level has distinct entries, and put \[S_k=\sum_{x\in A_k}\frac1x,\qquad B=\sum_{i=1}^r\frac{b_i}{a_i^2}.\] The level contains exactly $r^{k-1}$ distinct positive integers. Ordering them increasingly, the $j$th is at least $j$, so \[0<S_k\le H_{r^{k-1}}\le1+(k-1)\log r.\tag{1}\] Since all $a_i,b_i\ge1$, the minimum entry increases by at least one at each step. Thus every entry of $A_k$ is at least $k$.

The defining multiset recurrence, before any assumption of distinctness, gives \[S_{k+1}=\sum_{x\in A_k}\sum_i\frac1{a_ix+b_i}. \] The exact identity \[\frac1{a_ix+b_i}=\frac1{a_ix}-\frac{b_i}{a_ix(a_ix+b_i)} \ge\frac1{a_ix}-\frac{b_i}{a_i^2x^2} \] therefore implies \[S_{k+1}\ge C S_k-B\sum_{x\in A_k}\frac1{x^2} \ge\left(C-\frac Bk\right)S_k.\tag{2}\] Choose a fixed integer $K>B/(C-1)$ and set $\rho=C-B/K>1$. For every $k\ge K$, $2$ gives $S_{k+1}\ge\rho S_k$, and consequently \[S_k\ge\rho^{k-K}S_K.\tag{3}\] The positive constant $S_K$ makes $3$ exponential in $k$, contradicting $1$. Some level must contain a repeated integer, proving the assertion.

The strict reciprocal threshold is meaningful

For comparison, the two maps $x\mapsto2x+1$ and $x\mapsto2x+2$ have reciprocal sum exactly one. At depth $d$, their values at $1$ are \[2^d+\sum_{j=0}^{d-1}2^j c_j,\qquad c_j\in\{1,2\}. \] After subtracting $\sum_{j=0}^{d-1}2^j$, the remaining digits $c_j-1$ are binary. Different words therefore yield different values at the same depth. Thus replacing $>1$ by $\ge1$ would make the statement false.

Outcome and attribution

SOLVED. All estimates and the contradiction are elementary and included. The proof is the harmonic-weight method written by Kevin Barreto in the primary discussion, related there to Klarner's earlier affine-semigroup theorem; it is not a new solution. The stronger equality-of-affine-maps assertion is not required or claimed here. Sources: https://www.erdosproblems.com/481 and https://www.erdosproblems.com/forum/discuss/481, checked 24 September 2026. The existing Lean result belongs to its cited authors, and no local formal check of this text is claimed.

COMPLETION ESTIMATE

COMPLETION: 100%

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