gpt pro 5.4 attempt on Math-780

Round 2 continuation for Erdős problems #319, #359, #361

This file is written as a continuation of the Round-1 write-up. I explicitly indicate below whenever I use a Round-1 result as a black box.

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Problem #359

1) ROUND-2 OBJECTIVE

Path (C): obstruction/correction. The Round-1 write-up still had only the trivial lower bound $a_k\ge k$. The most natural strict advance is therefore to prove a genuinely stronger lower bound on $a_k$, or equivalently an upper bound on the counting function \[ A(x):=\#\{k:\ a_k\le x\}. \] I will prove the new unconditional estimate \[ A(x)\le \frac{3x+3}{4}, \qquad\text{hence}\qquad a_k\ge \frac{4k-3}{3}. \] This does not settle Erdős' question $a_k/k\to\infty$, but it is a rigorous improvement over Round 1 and it rules out any positive upper density larger than $3/4$.

2) ROUND-1 FOUNDATION USED

I use the following Round-1 points as fixed background.

• The original statement for general $n>1$ is internally inconsistent unless one inserts the monotonicity correction "least integer $>a_i$"; Round 1 already made this minimal correction.
• For the main asymptotic question, the case of real interest is still $n=1$.
• Round 1 proved the unconditional upper bound \[ a_{k+1}\le \binom{k+1}{2}+1. \]

In the present round I work only with the corrected $n=1$ sequence.

3) NEW INSIGHT / TOOL (ROUND-2)

The new observation is that adjacent two-term sums already force many permanent exclusions.

If \[ a_1<a_2<\cdots<a_t\le x/2, \] then the numbers \[ a_1+a_2,\ a_2+a_3,\ \dots,\ a_{t-1}+a_t \] are all distinct, all at most $x$, and none of them can ever appear later in the sequence, because each is already a sum of two consecutive earlier terms.

This gives a clean counting obstruction on $[1,x]$.

4) ATTACK PLAN (ROUND-2)

The precise gap after Round 1 was the absence of any nontrivial lower bound on $a_k$ beyond $a_k\ge k$.

To improve that, I will prove:

• (i) for every $x\ge 1$, \[ A(x)\le \frac{x+\lceil x/2\rceil+1}{2}\le \frac{3x+3}{4}; \]
• (ii) equivalently, \[ a_k\ge \left\lceil \frac{4k-3}{3}\right\rceil. \]

The mechanism is a direct "included numbers versus permanently forbidden adjacent sums" comparison inside the same interval $[1,x]$.

5) WORK (ROUND-2)

For the rest of this section let \[ A(x):=\#\{k:\ a_k\le x\} \] for the corrected $n=1$ MacMahon sequence.

Lemma 359.1.

For every integer $x\ge 1$, \[ A(x)\le \frac{x+\lceil x/2\rceil+1}{2}\le \frac{3x+3}{4}. \]

Proof.

Let \[ t:=A\!\left(\left\lfloor \frac{x}{2}\right\rfloor\right). \] Thus \[ a_1<a_2<\cdots<a_t\le \left\lfloor \frac{x}{2}\right\rfloor. \]

Consider the $t-1$ adjacent sums \[ s_i:=a_i+a_{i+1}\qquad (1\le i\le t-1). \] Because the sequence is strictly increasing, these sums are strictly increasing: \[ s_1<s_2<\cdots<s_{t-1}. \] Also \[ s_i\le 2\left\lfloor \frac{x}{2}\right\rfloor\le x. \] Finally, each $s_i$ is not a term of the sequence, because as soon as $a_{i+1}$ has appeared, the number $a_i+a_{i+1}$ is already representable as a sum of consecutive earlier terms, so it can never be chosen later by the defining greedy rule.

Hence, among the integers in $[1,x]$, there are at least $t-1$ numbers which are definitely not in the sequence. Therefore \[ A(x)\le x-(t-1)=x-t+1. \] On the other hand, from $\lfloor x/2\rfloor+1$ up to $x$ there are only \[ x-\left\lfloor \frac{x}{2}\right\rfloor=\left\lceil \frac{x}{2}\right\rceil \] available integers, so \[ A(x)-t\le \left\lceil \frac{x}{2}\right\rceil, \qquad\text{that is,}\qquad t\ge A(x)-\left\lceil \frac{x}{2}\right\rceil. \] Substituting this lower bound for $t$ into $A(x)\le x-t+1$ gives \[ A(x)\le x-\left(A(x)-\left\lceil \frac{x}{2}\right\rceil\right)+1, \] hence \[ 2A(x)\le x+\left\lceil \frac{x}{2}\right\rceil+1. \] Therefore \[ A(x)\le \frac{x+\lceil x/2\rceil+1}{2}\le \frac{3x+3}{4}. \] $\square$

Corollary 359.2.

For every $k\ge 1$, \[ a_k\ge \left\lceil \frac{4k-3}{3}\right\rceil. \]

Proof.

Apply Lemma 359.1 with $x=a_k$. Since $A(a_k)=k$, we obtain \[ k\le \frac{3a_k+3}{4}, \] so \[ a_k\ge \frac{4k-3}{3}. \] As $a_k$ is an integer, this yields \[ a_k\ge \left\lceil \frac{4k-3}{3}\right\rceil. \] $\square$

Consequence.

The counting function of the sequence has upper density at most $3/4$: \[ \limsup_{x\to\infty}\frac{A(x)}{x}\le \frac34. \] Equivalently, \[ \liminf_{k\to\infty}\frac{a_k}{k}\ge \frac43. \]

6) ADVERSARIAL VERIFICATION

I check the new proof against the main possible objections.

• Could two adjacent sums coincide? No. Since the sequence is strictly increasing, \[ a_i+a_{i+1}<a_{i+1}+a_{i+2} \] for every valid $i$.

• Could an adjacent sum later re-enter the sequence? No. Once $a_i$ and $a_{i+1}$ have been chosen, the value $a_i+a_{i+1}$ is permanently representable as a sum of consecutive earlier terms, so the greedy rule can never choose it later.

• Did the proof use anything false for small $x$? No. If $t=0$ or $t=1$, the argument still gives a valid inequality because there are then $t-1\le 0$ adjacent sums, and the final bound remains true. In particular, $A(1)=1$ and the estimate gives $1\le (3+1)/4$.

• Is the result genuinely stronger than Round 1? Yes. Round 1 had only $a_k\ge k$. We now have the explicit improvement \[ a_k\ge \frac43k+O(1). \]

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The open asymptotic questions from Erdős remain open, but Round 2 adds a real theorem that was not present in Round 1: \[ A(x)\le \frac{3x+3}{4} \qquad\text{and}\qquad a_k\ge \left\lceil \frac{4k-3}{3}\right\rceil. \] So the sequence is now known to have upper density at most $3/4$, which is a genuine structural obstruction beyond the trivial $a_k\ge k$.

8) COMPLETION ESTIMATE

COMPLETION: 35%

9) REFERENCES

• T. F. Bloom, Erdős Problem #359, erdosproblems.com/359 (checked in this round).
• Š. Porubský, On MacMahon's segmented numbers and related sequences, Nieuw Arch. Wisk. $3$ 25 $1977$, 403--408.

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