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Math-5386-A4 · revision r1 · on Math-5386

gpt pro 5.2 attempt on Math-5386, version 2

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Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: proved. Claims here do not change the source status.

Manuscript

Contents
  1. Erdős Problem #38 --- Round 2 (continuation & gap-closure)
  2. 1) ROUND-2 OBJECTIVE
  3. 2) ROUND-1 FOUNDATION USED
  4. 3) NEW INSIGHT / TOOL (ROUND-2)
  5. 4) ATTACK PLAN (ROUND-2)
  6. 5) WORK (ROUND-2)
  7. 5.1 A basic correction: sets missing $1$ have $d_s=0$
  8. 5.2 Corrected gcd obstruction (fixing Round 1 Lemma 38.1)
  9. 5.3 All-shifts union formula for periodic block sets when $\alpha\le 1/2$
  10. 5.4 Immediate forcing of $1\in B$
  11. 5.5 Bounded multiplicative gaps forced by $\alpha=1/2$
  12. 5.6 A small generalisation for rational $\alpha\le 1/2$
  13. 6) ADVERSARIAL VERIFICATION
  14. 7) FINAL
  15. 8) COMPLETION ESTIMATE (MANDATORY)
  16. 9) REFERENCES

Erdős Problem #38 --- Round 2 (continuation & gap-closure)

1) ROUND-2 OBJECTIVE

Path (C): obstruction/correction. Round 1 isolated one explicit obstruction (a gcd obstruction) and one adversarial family of sets $A$ (periodic block sets) but left two concrete gaps: (i) the gcd obstruction was built on a set $A=d\mathbb N$ whose Schnirelmann density is actually $0$ under the stated convention $\mathbb N=\{1,2,\dots\}$, (ii) the periodic-block calculation was only carried out for "small" shifts $b\le P-L$.

In Round 2 I (a) correct the gcd obstruction with a density-$1/d$ set that does contain $1$, and (b) strengthen the periodic-block calculation to an all-shifts formula (for $\alpha\le 1/2$) on a single period. This yields new necessary structural conditions on any candidate $B$, in particular that $1\in B$ and that $B$ cannot have unbounded multiplicative gaps.

2) ROUND-1 FOUNDATION USED

I rely on the following Round 1 components (without re-proving them):
  • Round 1 formal restatement of the problem and notation.
  • Round 1 Lemma 38.2 (periodic block sets): for $A_{P,L}=\bigcup_{k\ge 0}\{kP+1,\dots,kP+L\}$ one has $d_s(A_{P,L})=L/P$.
  • Round 1 Lemma 38.2 (finite-$N$ upper bound) in the regime $N=KP$ and $1\le b\le P-L$, giving \[ |(A_{P,L}\cup(A_{P,L}+b))\cap[1,N]|\le \left(\frac LP+\frac bP\right)N. \]
I will explicitly flag where Round 1 contained an error and supply a replacement argument.

3) NEW INSIGHT / TOOL (ROUND-2)

  • Schnirelmann sensitivity at $1$. Under $\mathbb N=\{1,2,\dots\}$, any $A$ with $1\notin A$ has $d_s(A)=0$. This invalidates the Round 1 use of $A=d\mathbb N$ as a positive-density example.
  • All-shifts union formula for block sets when $\alpha\le \tfrac12$. For $A_{P,L}$ with $L\le P/2$ and for every shift $0\le b\le P-1$, one can compute $|(A_{P,L}\cup(A_{P,L}+b))\cap[1,P]|$ exactly and deduce the sharp upper bound \[ \frac{|(A_{P,L}\cup(A_{P,L}+b))\cap[1,P]|}{P}\le \alpha+\frac{\min(b,P-b)}{P},\qquad \alpha=L/P\le \tfrac12. \]
  • New structural consequences for $B$. Specialising to $\alpha=1/2$ forces $1\in B$ (from $N=2$) and forces $B$ to intersect the "middle" interval $[f(1/2)P,\,P-f(1/2)P]$ for every even $P$, implying bounded multiplicative gaps $b_{n+1}/b_n\ll 1/f(1/2)$.

4) ATTACK PLAN (ROUND-2)

  1. Fix the Round 1 gcd obstruction. Prove the elementary lemma "$1\notin A\Rightarrow d_s(A)=0$" and replace $A=d\mathbb N$ by $A=\{1\}\cup d\mathbb N$, which has $d_s(A)=1/d$.
  2. Close the wrap-around gap for periodic blocks. Compute the union size on a single period $[1,P]$ for all shifts $0\le b<P$ when $L\le P/2$.
  3. Extract new necessary conditions on $B$ from $\alpha=1/2$. Apply the problem's property to $A_{P,P/2}$ at $N=P$ to force existence of a shift $b\in B\cap[1,P]$ with $\min(b,P-b)\ge f(1/2)P$. This yields: (i) $1\in B$ (take $P=2$), (ii) bounded multiplicative gaps (take $P$ just below $b_{n+1}$).
These steps strictly strengthen the Round 1 obstruction analysis without assuming anything about additive bases beyond the problem statement.

5) WORK (ROUND-2)

5.1 A basic correction: sets missing $1$ have $d_s=0$

Lemma (Sensitivity at $1$): If $A\subseteq\mathbb N$ and $1\notin A$, then $d_s(A)=0$.
Proof: By definition, \[ d_s(A)=\inf_{N\ge 1}\frac{|A\cap\{1,\dots,N\}|}{N}\le \frac{|A\cap\{1\}|}{1}=0. \] ∎

5.2 Corrected gcd obstruction (fixing Round 1 Lemma 38.1)

Round 1 argued with $A=d\mathbb N$, but Lemma lem:missing1 shows $d_s(d\mathbb N)=0$ under the present convention. The obstruction still holds after a minimal repair.
Lemma (Corrected gcd obstruction): Let $d\ge 2$. Suppose $B\subseteq d\mathbb N$. Then the desired property cannot hold (for any choice of $f$) because it fails already for $\alpha=1/d$. In particular, any candidate $B$ must satisfy $\gcd(B)=1$.
Proof: Define \[ A:=\{1\}\cup d\mathbb N=\{1,d,2d,3d,\dots\}. \] Claim 1: $d_s(A)=1/d$. For $N\ge 1$ we have $|A\cap[1,N]|=1+\lfloor N/d\rfloor$. Using $\lfloor N/d\rfloor\ge N/d-1$ gives \[ \frac{|A\cap[1,N]|}{N}=\frac{1+\lfloor N/d\rfloor}{N}\ge \frac{1+(N/d-1)}{N}=\frac1d. \] On the other hand for multiples $N=kd$ one has \[ \frac{|A\cap[1,kd]|}{kd}=\frac{1+k}{kd}=\frac1d+\frac{1}{kd}, \] which tends to $1/d$ as $k\to\infty$. Hence $d_s(A)=1/d$.

Now fix any $b\in B$. By assumption $b\equiv 0\pmod d$. Then $d\mathbb N+b\subseteq d\mathbb N$, so \[ A+b=(\{1\}+b)\cup(d\mathbb N+b)\subseteq \{1+b\}\cup d\mathbb N. \] Therefore \[ A\cup(A+b)=A\cup\{1+b\}. \] Consequently for every $N$ we have \[ |(A\cup(A+b))\cap[1,N]|\le |A\cap[1,N]|+1. \] Choose $N$ large so that $\frac{2}{N}<f(1/d)$. Since $\lfloor N/d\rfloor\le N/d$, we have \[ \frac{|A\cap[1,N]|}{N}=\frac{1+\lfloor N/d\rfloor}{N}\le \frac{1+N/d}{N}=\frac1d+\frac1N. \] Hence for every $b\in B$, \[ \frac{|(A\cup(A+b))\cap[1,N]|}{N}\le \frac{|A\cap[1,N]|}{N}+\frac1N\le \frac1d+\frac{2}{N}<\frac1d+f(1/d), \] contradicting the required inequality at $\alpha=1/d$. This proves the failure and hence forces $\gcd(B)=1$. ∎

5.3 All-shifts union formula for periodic block sets when $\alpha\le 1/2$

We now strengthen the Round 1 periodic-block computation by removing the restriction $b\le P-L$ when $L\le P/2$ and by working at a single period ($N=P$).
Lemma (One-period union formula for $\alpha\le 1/2$): Let integers $P\ge 2$ and $L$ satisfy $1\le L\le P/2$ and define the periodic block set \[ A_{P,L}:=\bigcup_{k\ge 0}\{kP+1,\dots,kP+L\}. \] Fix $b\in\{0,1,\dots,P-1\}$. Then \[ |(A_{P,L}\cup(A_{P,L}+b))\cap[1,P]|=L+\min\{L,\,b,\,P-b\}. \] In particular, writing $\alpha=L/P\le 1/2$, for every $1\le b\le P-1$ we have the sharp upper bound \[ \frac{|(A_{P,L}\cup(A_{P,L}+b))\cap[1,P]|}{P}\le \alpha+\frac{\min(b,P-b)}{P}. \]
Proof: On the first period we have $A_{P,L}\cap[1,P]=\{1,2,\dots,L\}$. For $0\le b\le P-1$, \[ (A_{P,L}+b)\cap[1,P]=\{a+b\le P: a\in A_{P,L}\}. \] Since $b\ge 0$, no element of $A_{P,L}$ from periods $\ge 1$ can shift into $[1,P]$. Thus only $a\in A_{P,L}\cap[1,P-b]=\{1,\dots,\min(L,P-b)\}$ contribute, and hence \[ (A_{P,L}+b)\cap[1,P]=\{b+1,\dots, b+\min(L,P-b)\}. \] Set $m:=\min(L,P-b)$; then $(A_{P,L}+b)\cap[1,P]$ is the interval $[b+1,b+m]$. We consider three cases.

Case 1: $0\le b<L$. Because $L\le P/2$, we have $P-b> P/2\ge L$, so $m=L$. The overlap of $[1,L]$ and $[b+1,b+L]$ is $[b+1,L]$, of size $L-b$. Therefore \[ |(A_{P,L}\cup(A_{P,L}+b))\cap[1,P]|=L+L-(L-b)=L+b. \] Since here $b<L$ and $P-b\ge L$, we have $\min\{L,b,P-b\}=b$, giving the claimed formula.

Case 2: $L\le b\le P-L$. Then $P-b\ge L$, hence $m=L$, and the shifted interval $[b+1,b+L]$ lies entirely to the right of $[1,L]$. So the two intervals are disjoint and \[ |(A_{P,L}\cup(A_{P,L}+b))\cap[1,P]|=L+L=2L. \] Here $\min\{L,b,P-b\}=L$, again matching the formula.

Case 3: $P-L<b\le P-1$. Then $P-b<L$, so $m=P-b$. Also $b> P-L\ge P/2\ge L$, so $b\ge L$ and the two intervals $[1,L]$ and $[b+1,P]$ are disjoint. Hence \[ |(A_{P,L}\cup(A_{P,L}+b))\cap[1,P]|=L+(P-b)=L+\min\{L,b,P-b\}, \] since in this regime $\min\{L,b,P-b\}=P-b$.

This exhausts all $b\in\{0,\dots,P-1\}$. The inequality follows immediately by dividing by $P$ and using $\min\{L,b,P-b\}\le \min\{b,P-b\}$. ∎

5.4 Immediate forcing of $1\in B$

Proposition: If a set $B$ satisfies the problem property for some function $f:(0,1)\to(0,\infty)$, then $1\in B$.
Proof: Take $\alpha=1/2$ and let $A$ be the odd numbers: $A=\{1,3,5,\dots\}$. Then $d_s(A)=1/2$ and for $N=2$ we have $A\cap[1,2]=\{1\}$. If $b\ge 2$ then $(A+b)\cap[1,2]=\emptyset$, so $|(A\cup(A+b))\cap[1,2]|=1$. But the required inequality at $\alpha=1/2$ is \[ |(A\cup(A+b))\cap[1,2]|\ge (1/2+f(1/2))\cdot 2 = 1+2f(1/2) > 1, \] forcing $|(A\cup(A+b))\cap[1,2]|\ge 2$, hence $b=1$. Thus $1\in B$. ∎

5.5 Bounded multiplicative gaps forced by $\alpha=1/2$

Set $f_0:=f(1/2)>0$.
Proposition (Middle-interval condition at every even scale): Assume $B$ satisfies the problem property with function $f$. Then for every even integer $P\ge 2$ there exists $b\in B\cap\{1,\dots,P-1\}$ such that \[ \min\{b,\,P-b\}\ge f_0 P. \] Equivalently, \[ B\cap [\,f_0 P,\,P-f_0 P\,]\neq\emptyset\qquad\text{for every even }P\ge 2. \]
Proof: Fix even $P\ge 2$ and consider the block set $A_{P,P/2}$. By Round 1 Lemma 38.2, $d_s(A_{P,P/2})=1/2$. Apply the problem property to this $A$ and $N=P$. Since shifting by $b\ge P$ produces no elements in $[1,P]$, the witnessing $b\in B$ must satisfy $1\le b\le P-1$. Lemma lem:allshifts (with $L=P/2$) gives \[ \frac{|(A\cup(A+b))\cap[1,P]|}{P}\le \frac12+\frac{\min(b,P-b)}{P}. \] The desired lower bound is $\frac12+f_0$. Therefore $\min(b,P-b)\ge f_0P$, as required. ∎
Corollary (No unbounded multiplicative gaps): Let $B=\{b_1<b_2<\cdots\}$. If $B$ satisfies the problem property, then \[ b_{n+1}\le \frac{1}{f_0}\,b_n+2\qquad\text{for all }n\ge 1. \] In particular $\limsup_{n\to\infty} b_{n+1}/b_n\le 1/f_0<\infty$, so $B$ cannot be lacunary.
Proof: Fix $n\ge 1$. If $b_{n+1}\le 2$ then $b_{n+1}\le 2\le b_n/f_0+2$ (since $b_n\ge 1$), so the claim is trivial. Thus we may assume $b_{n+1}\ge 3$, so the even integer $P$ defined below satisfies $P\ge 2$. Let $P$ be the largest even integer strictly smaller than $b_{n+1}$, i.e. $P=b_{n+1}-1$ if $b_{n+1}$ is odd and $P=b_{n+1}-2$ if $b_{n+1}$ is even. Then $P\ge b_{n+1}-2$ and $P<b_{n+1}$. Hence $B\cap[1,P]=\{b_1,\dots,b_n\}$. By Proposition Proposition (Middle-interval condition at every even scale) there exists $b\in B\cap[1,P]$ with $b\ge f_0P$. Since $b\le b_n$ we get $b_n\ge f_0P\ge f_0(b_{n+1}-2)$. Rearranging yields $b_{n+1}\le b_n/f_0+2$.

The ratio statement follows immediately. ∎

5.6 A small generalisation for rational $\alpha\le 1/2$

For later use it is convenient to record the same "middle interval" forcing for any rational density $\alpha\le 1/2$.
Proposition (Middle-interval condition for rational $\alpha\le 1/2$): Fix $\alpha\in(0,1/2]$ and write $f_\alpha:=f(\alpha)>0$. Assume $\alpha=L_0/P_0$ in lowest terms. Then for every multiple $P=mP_0$ (with $m\in\mathbb N$) there exists $b\in B\cap\{1,\dots,P-1\}$ such that \[ \min\{b,\,P-b\}\ge f_\alpha P. \] Equivalently, $B\cap[f_\alpha P,\,P-f_\alpha P]\neq\emptyset$ for every such $P$.
Proof: Take $A=A_{P,L}$ with $L=\alpha P=mL_0$ so that $d_s(A)=\alpha$ by Round 1 Lemma 38.2. Apply the problem property at $N=P$. As before, the witnessing $b$ must satisfy $1\le b\le P-1$. Lemma lem:allshifts gives \[ \frac{|(A\cup(A+b))\cap[1,P]|}{P}\le \alpha+\frac{\min(b,P-b)}{P}. \] Comparing with the required lower bound $\alpha+f_\alpha$ forces $\min(b,P-b)\ge f_\alpha P$. ∎

6) ADVERSARIAL VERIFICATION

  • Check the corrected gcd obstruction. The only delicate point is that $d\mathbb N+b\subseteq d\mathbb N$ (not equality), but this suffices because it implies $A\cup(A+b)=A\cup\{1+b\}$ and hence the union count increases by at most $1$. The computed Schnirelmann density $d_s(\{1\}\cup d\mathbb N)=1/d$ is robust because the ratio is bounded below by $1/d$ for all $N$ and approaches $1/d$ along $N=kd$.
  • Quantifiers / "for every $N$" issues. All deductions from $\alpha=1/2$ used the problem requirement at specific even values $N=P$ (not asymptotic $N\to\infty$), so there is no hidden passage to a subsequence.
  • Boundary cases in Lemma lem:allshifts. The casework uses $L\le P/2$ to ensure in Case 1 that $P-b\ge L$ and in Case 3 that $b\ge L$; these are exactly where wrap-around could otherwise appear. The endpoints $b=0$ and $b=P-1$ are covered: the formula gives union size $L$ and $L+1$ respectively, matching the direct interval picture.
  • Could Proposition Proposition (Middle-interval condition at every even scale) fail because the witness $b$ might be $\ge P$? No: if $b\ge P$ then $(A_{P,P/2}+b)\cap[1,P]=\emptyset$, so the union count is $P/2$, which cannot reach $(1/2+f_0)P$ since $f_0>0$.
  • Internal consistency with Round 1. Round 1 Lemma 38.2 remains correct (it crucially uses sets containing $1$), and Lemma lem:allshifts is a strict strengthening on the special case $N=P$ and $\alpha\le 1/2$. The only Round 1 piece that needed correction was the use of $A=d\mathbb N$ as a positive-$d_s$ example.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED). We have not resolved whether such a non-basis $B$ exists. However, Round 2 makes two substantive advances beyond Round 1:
  1. The gcd obstruction is repaired in a way compatible with the stated Schnirelmann density convention (Lemma Lemma (Corrected gcd obstruction)).
  2. New necessary structural conditions are proved for any candidate $B$:
These conditions rule out broad classes of natural non-bases (e.g. lacunary sequences) and tighten the search space for a potential counterexample.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 35%.

9) REFERENCES

No external references were used beyond the problem statement and Round 1 results.