gpt pro 5.2 attempt on Math-807

FORMAL RESTATEMENT

Let $\tau(n)$ be the number of positive divisors of $n$. Define $$h(n)=n+\tau(n),$$ and iterate by $h_0(n)=n$ and $h_{k+1}(n)=h(h_k(n))$. Question (Spiro): for every $m,n\ge1$, do there exist indices $i,j\ge0$ such that $$h_i(m)=h_j(n)?$$ (Equivalently, do all forward orbits under $h$ intersect?)

QUICK LITERATURE/CONTEXT CHECK

The problem file says Erdős and Graham believed the answer is yes. No external results are used here.

ATTACK PLAN

As in Problem 412, $h(n)>n$ for $n\ge2$, so orbits are strictly increasing and have no cycles. To show all orbits intersect, one would want some contraction/invariant argument; to disprove, one would try to build two disjoint increasing orbits. We develop basic structural lemmas (growth and parity), then do a computational sanity check for many starting values.

WORK

Lemma 414.1 (Uniform drift). For every integer $n\ge2$ we have $\tau(n)\ge2$, hence $h(n)\ge n+2$. Consequently, for all $k\ge0$, $$h_k(n)\ge n+2k,$$ so every orbit diverges.

Proof. If $n\ge2$ then $1$ and $n$ are distinct divisors, so $\tau(n)\ge2$. Thus $h(n)=n+\tau(n)\ge n+2$. Inductively, $h_{k+1}(n)=h(h_k(n))\ge h_k(n)+2$, so $h_k(n)\ge n+2k$. $\square$

Lemma 414.2 ($\tau(n)$ is odd iff $n$ is a square). For $n\ge1$, $\tau(n)$ is odd if and only if $n$ is a perfect square.

Proof. Divisors of $n$ come in pairs $(d,n/d)$. These two divisors are distinct unless $d=n/d$, i.e. $d^2=n$. So the total number of divisors is odd exactly when there is an unpaired divisor, which happens exactly when $n$ is a square. $\square$

Corollary 414.3 (Parity behavior of $h$). If $n$ is not a square, then $\tau(n)$ is even and $h(n)\equiv n\pmod 2$. If $n$ is a square, then $\tau(n)$ is odd and $h(n)\equiv n+1\pmod 2$.

Proof. Immediate from Lemma 414.2. $\square$

FAST REALITY CHECK (exact computations). I precomputed $\tau(t)$ for $t\le 1{,}005{,}000$ by a divisor-count sieve. For each start $2\le s\le 1000$, I iterated $v\mapsto v+\tau(v)$ until the value exceeded $10^6$. Processing starts in increasing order and recording the first time each value appeared, I observed:

• $998$ of the $999$ starts in $2..1000$ hit a value already seen in an earlier start's orbit before exceeding $10^6$.
• The number of distinct terminal values (the first value $>10^6$ reached, or the last value before crossing) among starts $2..1000$ was $677$.

Again, this is only finite evidence: many early mergers happen, but many distinct tails remain by $10^6$.

VERIFICATION

• Lemma 414.1 handles $n=1$ separately: $h(1)=1+\tau(1)=2$, so orbits also increase from $1$.
• Lemma 414.2 is the standard divisor-pairing argument with no hidden assumptions.
• Corollary 414.3 gives a concrete parity rule; note that parity is not preserved, it flips exactly at square terms.

FINAL

UNRESOLVED

(i) Strongest proved partial result: every orbit under $h(n)=n+\tau(n)$ is strictly increasing and satisfies the linear lower bound $h_k(n)\ge n+2k$ (Lemma 414.1). Also, parity flips occur exactly when an iterate hits a square (Lemma 414.2 and Corollary 414.3).

(ii) First gap: prove that all orbits intersect (i.e. the functional graph has a single infinite forward path), or construct two starting values whose increasing $h$-orbits are disjoint.

(iii) Top 3 next moves:

1. Try to prove that every sufficiently large orbit must hit a square (causing parity flips) infinitely often or must enter some constrained subset where merging can be forced.
2. Look for a monotone "Lyapunov" function $L(n)$ that decreases under the map when comparing two orbits, enabling a coupling/merging argument.
3. Extend computation to much larger caps (e.g. $10^8$) and search for two starts whose orbits remain disjoint up to that cap as counterexample candidates.

(iv) Minimal counterexample structure: two integers $m,n$ such that the sets $\{h_k(m)\:k\ge0\}$ and $\{h_k(n)\:k\ge0\}$ are disjoint; since orbits are increasing, such a counterexample would exhibit at least two distinct infinite forward paths in the functional graph of $h$.