gpt pro 5.4 attempt on Math-692

Erdős Problem #148 (Round 2): a self-contained lower bound of size $\exp(\Omega(k^2))$

1) ROUND-2 OBJECTIVE

For Problem #148, the original task is not a single true/false statement but an estimation problem. So paths (A) and (B) from the generic Round-2 template do not literally apply. I therefore pursue the Round-2 analogue of path (C): I keep the statement as an asymptotic estimation problem and strengthen the lower-bound side of the estimate.

Concretely, Round 1 produced the fully self-contained bound \[ F(k)\ge 3^{k-3}. \] The exact Round-2 goal is to prove a substantially stronger self-contained lower bound.

2) Round-1 FOUNDATION USED

I use the following Round-1 ingredients and do not reprove them from scratch.

1. The Round-1 splitting identity: \[ \frac1N=\frac1{N+m}+\frac1{N(N+m)/m} \qquad\mbox{whenever }m\mid N,\ 1\le m<N. \]
2. The Round-1 strategy of always splitting the current largest denominator.
3. The Round-1 observation that if one always splits the current largest denominator, then all smaller denominators created in earlier steps remain untouched forever.

3) NEW INSIGHT / TOOL (ROUND 2)

The new ingredient is a two-phase branching scheme.

1. Divisor-amplification phase. Repeatedly use the choice $m=1$ to force the current largest denominator to acquire many divisors.
2. Frozen-divisor branching phase. Once a divisor-rich denominator $M$ has been created, use every proper divisor of $M$ as a branching choice at every later step. Because every later largest denominator remains a multiple of $M$, all these choices remain valid forever.

This replaces the Round-1 fixed three-choice branching argument by a branching factor that grows exponentially during the build-up phase.

4) ATTACK PLAN (ROUND 2)

The exact gap after Round 1 was this: the lower bound came from keeping only the three divisors $1,2,3$ of the initial largest denominator $6$. That leaves a lot of unused structure, because the current largest denominator typically acquires many more divisors after repeated splittings.

So I will prove the following claims.

1. After $r$ repeated uses of $m=1$, the largest denominator $M_r$ has at least $2^{r+2}$ positive divisors.
2. Once the current largest denominator is divisible by $M_r$, every proper divisor of $M_r$ yields a valid next split, and this divisibility by $M_r$ is preserved forever.
3. Distinct branching sequences in the second phase yield distinct final Egyptian-fraction representations.

From these claims one gets a family of lower bounds \[ F(3+r+s)\ge (\tau(M_r)-1)^s, \] where $\tau(n)$ denotes the number of positive divisors of $n$. Optimising over $r+s=k-3$ will then give the explicit estimate \[ F(k)\ge 2^{\lfloor (k-3)^2/4\rfloor}. \]

5) WORK (ROUND 2)

Step 1: build-up by repeated choice $m=1$.

Start from the Round-1 base representation \[ 1=\frac12+\frac13+\frac16. \] Let $M_0:=6$. For $j\ge 0$, define recursively \[ M_{j+1}:=M_j(M_j+1), \] which is exactly the new largest denominator obtained from $M_j$ by the Round-1 splitting identity with $m=1$: \[ \frac1{M_j}=\frac1{M_j+1}+\frac1{M_j(M_j+1)}. \]

Thus, after $r$ such steps we obtain a valid representation of $1$ with $3+r$ terms whose largest denominator is $M_r$.

Step 2: divisor count grows at least exponentially.

Let $\tau(n)$ denote the number of positive divisors of $n$. Since consecutive integers are coprime, \[ \gcd(M_j,M_j+1)=1, \] hence \[ \tau(M_{j+1})=\tau(M_j)\tau(M_j+1)\ge 2\tau(M_j) \] because $M_j+1>1$ and therefore $\tau(M_j+1)\ge 2$.

Since $\tau(M_0)=\tau(6)=4$, induction gives \[ \tau(M_r)\ge 4\cdot 2^r=2^{r+2}. \]

Step 3: all proper divisors of $M_r$ stay available forever.

Fix $r\ge 0$, and suppose we are at some later stage of the process where the current largest denominator is $L$ and $M_r\mid L$.

Let $d$ be any proper divisor of $M_r$. Then $d\mid L$ as well, so the Round-1 splitting identity gives \[ \frac1L=\frac1{L+d}+\frac1{L(L+d)/d}. \] The new largest denominator is \[ L':=\frac{L(L+d)}{d}=L\left(\frac{L}{d}+1\right). \] Since $M_r\mid L$, it follows immediately that $M_r\mid L'$. Therefore the property \[ M_r \mid \mbox{(current largest denominator)} \] is preserved after every such split.

Consequently, once the process reaches $M_r$, there are always at least \[ \tau(M_r)-1 \] available branching choices at every later step, namely all proper divisors of $M_r$.

Step 4: injectivity of the branching phase.

Now fix $r\ge 0$. After the build-up phase we have a representation whose current largest denominator is $M_r$.

Run a second phase of length $s\ge 0$. At each of the $s$ steps, choose an arbitrary proper divisor $d_i$ of $M_r$ and split the current largest denominator $L_i$ using $d_i$: \[ \frac1{L_i}=\frac1{L_i+d_i}+\frac1{L_{i+1}}, \qquad L_{i+1}=\frac{L_i(L_i+d_i)}{d_i}. \]

I claim that distinct sequences $(d_1,\dots,d_s)$ produce distinct final Egyptian-fraction representations.

Indeed, at step $i$ the split creates the denominator \[ a_i:=L_i+d_i. \] Because $d_i<L_i$, one has \[ a_i=L_i+d_i < \frac{L_i(L_i+d_i)}{d_i}=L_{i+1}. \] So $a_i$ is not the new largest denominator. Since all later steps split only the current largest denominator, the denominator $a_i$ is never altered again. In particular, each $a_i$ survives to the final representation.

Now suppose two branching sequences first differ at time $t$. Up to step $t-1$ they have the same current largest denominator $L_t$, but then choose different proper divisors $d_t\neq d'_t$. Hence they create different surviving denominators \[ a_t=L_t+d_t \neq L_t+d'_t=a'_t. \] Since both $a_t$ and $a'_t$ persist to the end, the two final representations are different.

Therefore the number of distinct final representations produced by an $s$-step branching phase is at least \[ (\tau(M_r)-1)^s. \]

Step 5: the general lower bound.

Combining the previous steps, we have proved:

Theorem. For every pair of integers $r,s\ge 0$, \[ F(3+r+s)\ge (\tau(M_r)-1)^s, \] where \[ M_0=6,\qquad M_{j+1}=M_j(M_j+1). \] In particular, since $\tau(M_r)\ge 2^{r+2}$, \[ F(3+r+s)\ge (2^{r+2}-1)^s. \]

Now set \[ r:=\lfloor \frac{k-3}{2}\rfloor, \qquad s:=k-3-r=\lceil \frac{k-3}{2}\rceil. \] Then \[ F(k)\ge (2^{r+2}-1)^s \ge (2^r)^s = 2^{rs}. \] Since \[ rs=\lfloor \frac{(k-3)^2}{4}\rfloor, \] we obtain the explicit Round-2 lower bound \[ F(k)\ge 2^{\lfloor (k-3)^2/4\rfloor}. \]

Conclusion for Problem #148.

Round 2 therefore upgrades the Round-1 self-contained lower bound from \[ F(k)\ge 3^{k-3} \] to the much stronger estimate \[ F(k)\ge 2^{\lfloor (k-3)^2/4\rfloor}. \] This is a genuine quantitative improvement from merely exponential growth in $k$ to growth of order $\exp(\Omega(k^2))$.

6) ADVERSARIAL VERIFICATION

I check the main possible failure modes.

1. Could a chosen divisor fail to remain valid later? No. Every branching divisor $d$ is chosen from the fixed divisor set of $M_r$, and every later largest denominator remains a multiple of $M_r$, hence also a multiple of $d$.
2. Could two different branching sequences merge later? The revised injectivity argument avoids this issue entirely. It does not rely on the final largest denominator. Instead it uses the permanently surviving denominators $a_i=L_i+d_i$, which encode the first place where two sequences differ.
3. Could one of the survivor denominators be altered later? No. At the moment it is created, it is smaller than the new largest denominator $L_{i+1}$, and every later operation only replaces the current largest denominator.
4. Are the resulting denominators still distinct and increasing? Yes. Every newly created denominator is strictly larger than the previous largest denominator, hence larger than all pre-existing denominators.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 2 does not settle the full asymptotic growth problem, but it makes a real mathematical advance over Round 1: it proves the self-contained bound \[ F(k)\ge 2^{\lfloor (k-3)^2/4\rfloor}, \] which is much stronger than the Round-1 bound $3^{k-3}$.

The remaining gap is still enormous compared with the best known lower bound recorded on the Erdős Problems site, but the internal branching method is now meaningfully sharper.

8) COMPLETION ESTIMATE

COMPLETION: 55%

9) REFERENCES

1. Erdős Problems, Problem #148, "For each $k$, let $F(k)$ be the number of representations of $1$ as a sum of $k$ distinct unit fractions...", checked on 7 March 2026.