Erdős Problem #521

ROUND-2 OBJECTIVE

Erdős problem number: 521.

I pursue path (C) obstruction/correction. The previous round already isolated the main unresolved gap: one can prove almost sure asymptotics on suitable sparse subsequences, but there is still no maximal inequality strong enough to upgrade this to the full sequence for the outside roots. I do not have a gap-free proof or counterexample for the original statement.

What I will do in this round is strictly advance the state of the investigation by proving a new corrected statement that is stronger than convergence in probability and much denser than the deterministic subsequence laws from the previous round:

almost surely, the exceptional set of degrees $n$ for which $R_n/\log n$ deviates from $2/\pi$ has finite harmonic measure. In particular, almost surely there exists a random set $G\subset\mathbb N$ with $\sum_{n\notin G}1/n<\infty$ such that $R_n/\log n\to 2/\pi$ along $n\to\infty$, $n\in G$.

Round-(round_no-1) FOUNDATION USED

I use the following vetted results from the previous round.

1. Multiple-root negligibility. If $D_n$ denotes the event that $f_n$ has a real multiple root, then \[ \sum_{n=1}^{\infty}\mathbb P(D_n)<\infty. \] Hence, almost surely, every real root of $f_n$ is simple for all sufficiently large $n$. Therefore asymptotic statements for distinct real-root counts and multiplicity-counting real-root counts are equivalent.

2. Expectation constants. For the distinct-root counts $N_n(I)$, the previous round established \[ \mathbb E N_n([0,1]) =\mathbb E N_n([-1,0]) =\mathbb E N_n([1,\infty)) =\mathbb E N_n((-\infty,-1]) =\frac{1}{2\pi}\log n+o(\log n), \] and also \[ \mathbb E N_n(\mathbb R)=\frac{2}{\pi}\log n+O(1). \]

3. Endpoint formula. The previous round computed the exact probabilities \[ \mathbb P(f_n(1)=0)=\mathbb P(f_n(-1)=0)= \begin{cases} 0,& n\text{ even},\\[3pt] 2^{-(n+1)}\binom{n+1}{(n+1)/2},& n\text{ odd}, \end{cases} \] so in particular \[ \mathbb P(f_n(\pm1)=0)=O(n^{-1/2}). \]

4. Dense deterministic subsequence theorem. The previous round proved that if $(n_k)$ is deterministic and \[ \frac{\sqrt{\log n_k}}{\log k}\to\infty, \] then almost surely \[ \frac{R_{n_k}}{\log n_k}\to\frac{2}{\pi}, \qquad \frac{R_{n_k}([1,\infty))}{\log n_k}\to\frac{1}{2\pi}, \qquad \frac{R_{n_k}((-\infty,-1])}{\log n_k}\to\frac{1}{2\pi}. \] I will not re-prove this. The new theorem below is denser in a different direction.

NEW INSIGHT / TOOL (ROUND-2)

The new tool is a harmonic-weighted Borel--Cantelli argument.

The available concentration estimate has the form \[ \mathbb P\left(\left|N_n(I)-\mathbb E N_n(I)\right|\ge \varepsilon\log n\right) \le C_{\varepsilon,I}e^{-c_{\varepsilon,I}\sqrt{\log n}}, \] for the relevant intervals $I$. This bound is not summable in $n$, so it cannot prove the full almost sure convergence. However, after inserting the critical weight $1/n$, the tails do become summable: \[ \sum_{n\ge 3}\frac{1}{n}e^{-c\sqrt{\log n}}<\infty. \] This leads to the almost sure statement \[ \sum_{n\ge 3}\frac{1}{n} \mathbf 1_{\{\,|N_n(I)/\log n-c_I|>\varepsilon\,\}}<\infty, \] which is substantially stronger than convergence in probability. It says the bad degrees form a random set of finite harmonic sum.

This is genuinely new relative to the previous round.

ATTACK PLAN (ROUND-2)

The exact gap after the previous round is still the lack of a maximal inequality that controls oscillations between nearby degrees. I do not try to bridge that gap directly.

Instead I prove the following chain of rigorous claims.

1. A weighted summability lemma: \[ \sum_{n\ge3}\frac1n e^{-c\sqrt{\log n}}<\infty. \]

2. A weighted Tonelli--Borel--Cantelli lemma: if $\sum \mathbb P(E_n)/n<\infty$, then almost surely $\sum \mathbf 1_{E_n}/n<\infty$.

3. Using the previous round's expectation asymptotics and Can--Nguyen's concentration theorem, prove that for each fixed $\varepsilon>0$ and each relevant interval $I$, the bad set \[ \left\{n:\left|N_n(I)/\log n-c_I\right|>\varepsilon\right\} \] has almost surely finite harmonic sum.

4. From this, construct a random set $G$ with $\sum_{n\notin G}1/n<\infty$ along which the normalized root counts converge to the expected constants.

This strictly advances beyond the previous round because it gives a random set of full logarithmic density, rather than just a deterministic sparse subsequence.

WORK (ROUND-2)

For an interval $I\subseteq\mathbb R$, let $N_n(I)$ denote the number of distinct real roots of $f_n$ in $I$. Let \[ \mathcal I:=\{\mathbb R,[0,1],[-1,0],[1,\infty),(-\infty,-1]\}. \] For $I\in\mathcal I$, define \[ c_I:= \begin{cases} \dfrac{2}{\pi},& I=\mathbb R,\\[6pt] \dfrac{1}{2\pi},& I\in\{[0,1],[-1,0],[1,\infty),(-\infty,-1]\}. \end{cases} \] By the previous round, for every $I\in\mathcal I$, \begin{equation}\label{wsref-0-65-71-3a-65-78-70-65-63-74-2d-61-73-79} \mathbb E N_n(I)=c_I\log n+o(\log n). \end{equation}

Lemma 1 (weighted tail summability)

Proof.

Lemma 2 (Tonelli weighted Borel--Cantelli)

Proof.

Lemma 3 (weighted concentration around the asymptotic constant)

Proof.

Theorem 4 (harmonic-summable exceptional set for a fixed interval)

Proof.

Corollary 5 (endpoint-root degrees are harmonic-summable)

Proof.

Theorem 6 (logarithmic-density-one convergence for distinct root counts)

Proof.

Choose integers \[ 3\le M_1<M_2<\cdots \] such that for each $m\ge1$, \[ \sum_{\substack{n\in B_m(\omega)\\ n\ge M_m}}\frac1n<2^{-m}. \] Now define \[ G_I(\omega):= \bigcup_{m=1}^{\infty} \left([M_m,M_{m+1})\cap B_m(\omega)^c\right). \] If $n\in G_I(\omega)$ and $n\in[M_m,M_{m+1})$, then \[ \left|\frac{N_n(I)(\omega)}{\log n}-c_I\right|\le \frac1m. \] Hence (eq:GI-limit) holds.

It remains to prove (eq:harmonic-good). If $n\notin G_I(\omega)$ and $n\in[M_m,M_{m+1})$, then $n\in B_m(\omega)$. Therefore \[ \sum_{\substack{n\ge3\\ n\notin G_I(\omega)}}\frac1n \le \sum_{m=1}^{\infty} \sum_{\substack{n\in B_m(\omega)\\ n\ge M_m}}\frac1n <\sum_{m=1}^{\infty}2^{-m}=1. \] This proves (eq:harmonic-good). $\square$

Corollary 7 (simultaneous convergence on a set of full logarithmic density)

Proof.

Corollary 8 (the original multiplicity-counting problem holds on a set of full logarithmic density)

Proof.

Corollary 9 (bad degrees have logarithmic density $0$)

Proof.

ADVERSARIAL VERIFICATION

I now check the new work for possible failure points.

1. Did I really advance beyond the previous round? Yes. The previous round gave deterministic dense-subsequence laws. The present round gives a much denser random set $G$ with \[ \sum_{n\notin G}\frac1n<\infty, \] along which the conjectured asymptotics hold. This is stronger in density and is not implied by the previous subsequence theorem.

2. Did I accidentally use full-sequence convergence anywhere? No. Every proof uses only the previous round's expectation asymptotics, endpoint formula, multiple-root lemma, and the known one-point concentration estimate.

3. Is there any hidden independence assumption across $n$? No. Lemma 2 uses only Tonelli's theorem. No independence between the events $E_n$ is required.

4. What about the overlap of closed intervals at $\pm1$? This is handled explicitly in Corollary 7 by intersecting with $G_{\mathrm{ep}}$, the set on which $f_n(1)\neq0$ and $f_n(-1)\neq0$. Since the exceptional endpoint degrees have finite harmonic sum almost surely, removing them preserves the theorem.

5. What about multiplicities? They are harmless because the previous round proved that real multiple roots occur only finitely often almost surely. Corollary 8 uses this explicitly.

6. Does finite harmonic sum really imply the announced density statement? Yes. If $A\subseteq\mathbb N$ satisfies $\sum_{n\in A}1/n<\infty$, then automatically \[ \frac{1}{\log N}\sum_{\substack{n\le N\\ n\in A}}\frac1n\to0. \] This is exactly the logarithmic-density-zero conclusion for the bad degrees.

I do not see a gap in the new argument.

FINAL

UNRESOLVED (BUT STRICTLY ADVANCED)

The original full-sequence almost sure convergence problem remains open. However, this round proves the following new theorem.

Proved corrected statement. Almost surely there exists a random set of degrees $G\subseteq\mathbb N$ with \[ \sum_{n\notin G}\frac1n<\infty \] such that, as $n\to\infty$ along $n\in G$, \[ \frac{R_n}{\log n}\to\frac{2}{\pi}, \qquad \frac{R_n((-\infty,-1]\cup[1,\infty))}{\log n}\to\frac1\pi, \qquad \frac{R_n([1,\infty))}{\log n}\to\frac1{2\pi}, \qquad \frac{R_n((-\infty,-1])}{\log n}\to\frac1{2\pi}. \] In particular, for every fixed $\varepsilon>0$, the set of degrees \[ \left\{n\ge3:\left|\frac{R_n}{\log n}-\frac{2}{\pi}\right|>\varepsilon\right\} \] has almost surely finite harmonic sum.

Sharp obstruction isolated. The current one-point concentration estimate is not summable over all $n$, so Borel--Cantelli alone cannot prove the original statement. The new theorem shows that current methods do prove the conjectured asymptotic outside a random harmonic-summable exceptional set. What remains missing is precisely a method to upgrade this to the full sequence.

COMPLETION ESTIMATE

COMPLETION: 81%

REFERENCES

1. Y. Q. Do, A strong law of large numbers for real roots of random polynomials, arXiv:2403.06353.
2. V. H. Can and O. Nguyen, Concentration inequalities for the number of real zeros of Kac polynomials, arXiv:2311.15446.
3. Y. Q. Do, H. Nguyen, and V. Vu, Real roots of random polynomials: expectation and repulsion, Proc. Lond. Math. Soc. $3$ 111 $2015$, no. 6, 1231--1260.
4. T. F. Bloom, Erdős Problem #521, erdosproblems.com/521.

ROUND-2 OBJECTIVE

I pursue path (C) obstruction/correction. The Round-1 output already reduced the problem to a specific missing ingredient: a maximal inequality capable of upgrading subsequence convergence to full-sequence convergence for the outside roots. I do not have a gap-free proof or counterexample for the original full-sequence statement. Instead, I will do two things that strictly advance the investigation.

First, I isolate a precise obstruction: the currently available one-point concentration bound is not summable along the full sequence, so Borel--Cantelli alone can never prove the desired almost sure full-sequence law. Second, I push the Borel--Cantelli method to its natural limit and prove a substantially stronger subsequence theorem than Round 1: the almost sure asymptotic law holds not only on lacunary subsequences, but on every deterministic subsequence $(n_k)$ satisfying \[ \frac{\sqrt{\log n_k}}{\log k}\to\infty. \] This includes explicit non-lacunary subsequences with \[ \frac{n_{k+1}}{n_k}\to 1. \] Using this, together with Round-1 symmetry lemmas and Do's strong law on $[-1,1]$, I obtain exact almost sure asymptotics along such dense subsequences for the hard outside intervals $[1,\infty)$, $(-\infty,-1]$, and their union.

Round-1 FOUNDATION USED

I will use the following vetted Round-1 results without reproving them.

• Round-1 multiple-root lemma. If $D_n$ denotes the event that $f_n$ has a real multiple root, then \[ \sum_{n=1}^\infty \mathbb P(D_n)<\infty. \] Hence, almost surely, all real roots of $f_n$ are simple for all sufficiently large $n$. In particular, for all large $n$, counting real roots with multiplicity agrees with counting distinct real roots on every real interval.

• Round-1 endpoint formula. For each $n\ge 0$, \[ \mathbb P(f_n(1)=0)=\mathbb P(f_n(-1)=0)= \begin{cases} 0,& n \text{ even},\\[3pt] 2^{-(n+1)}\binom{n+1}{(n+1)/2},& n \text{ odd}. \end{cases} \] In particular, \[ \mathbb P(f_n(\pm 1)=0)=O(n^{-1/2}). \]

• Round-1 reciprocal expectation symmetry. If $N_n(a,b)$ denotes the number of distinct real roots of $f_n$ in $(a,b)$, then \[ \mathbb E N_n(0,1)=\mathbb E N_n(1,\infty), \qquad \mathbb E N_n(-1,0)=\mathbb E N_n(-\infty,-1). \]

• Round-1 lacunary global law. Along any lacunary subsequence $(n_k)$ with $n_{k+1}\ge q n_k$ for some fixed $q>1$, \[ \frac{R_{n_k}}{\log n_k}\to \frac{2}{\pi} \qquad\text{almost surely.} \] The present round strengthens this substantially.

NEW INSIGHT / TOOL (ROUND-2)

The new input is the following observation.

Summability, not lacunarity, is the real issue. For each fixed $\varepsilon>0$ and each interval \[ I\in\{\mathbb R,[0,1],[1,\infty),[-1,0],(-\infty,-1]\}, \] Can--Nguyen's concentration theorem (quoted explicitly in Do's paper) gives constants $c_{\varepsilon,I},C_{\varepsilon,I}>0$ such that \[ \mathbb P\left(|N_n(I)-\mathbb E N_n(I)|\ge \varepsilon \log n\right) \le C_{\varepsilon,I} e^{-c_{\varepsilon,I}\sqrt{\log n}} \qquad(n\ge 2). \] Therefore, Borel--Cantelli works on every deterministic subsequence $(n_k)$ for which these tails are summable. Lacunarity is only one sufficient condition, not the correct boundary.

A convenient explicit sufficient condition is \[ \frac{\sqrt{\log n_k}}{\log k}\to\infty. \] This still allows very dense subsequences, including ones with $n_{k+1}/n_k\to 1$.

Second new observation. The same summable-tail argument applies not only to the total real-root count, but also to each of the four fundamental real intervals separately. After identifying their expectation constants using the Round-1 symmetry lemmas, one obtains almost sure asymptotics for the outside half-lines $[1,\infty)$ and $(-\infty,-1]$ along these dense subsequences.

ATTACK PLAN (ROUND-2)

The exact Round-1 gap was this: the Borel--Cantelli argument gives almost sure convergence along sparse subsequences, but cannot by itself fill the gaps between neighboring degrees. The missing ingredient is a maximal inequality controlling \[ \max_{n\le m\le n(1+o(1))}|N_m-N_n|. \] Do proved such a maximal inequality for roots in $[0,1]$, which yields the full-sequence strong law there, but explicitly explained that his pairing argument does not extend to the reciprocal-polynomial process needed for the outside roots.

I therefore proceed in two steps.

1. Prove a rigorous obstruction lemma: the one-point tail bound $e^{-c\sqrt{\log n}}$ is not summable along the full sequence, so one-point concentration alone cannot prove the full-sequence strong law.

2. Prove a dense-subsequence strong law: along any deterministic subsequence with \[ \frac{\sqrt{\log n_k}}{\log k}\to\infty, \] we have almost sure convergence of $N_{n_k}(I)/\log n_k$ to the expected constant for each interval \[ I\in\{\mathbb R,[0,1],[-1,0],[1,\infty),(-\infty,-1]\}. \] This gives new a.s. laws for the outside roots and exhibits explicit non-lacunary subsequences with ratio $n_{k+1}/n_k\to1$ on which the conjectured constant is already forced.

This strictly advances beyond Round 1 because it replaces the lacunary theorem by a much denser subsequence theorem and, for the first time in this investigation, yields an exact almost sure law for the outside half-lines along explicit non-lacunary subsequences.

WORK (ROUND-2)

Throughout this section, $N_n(I)$ denotes the number of distinct real roots of $f_n$ in the interval $I$. By the Round-1 multiple-root lemma, for almost every outcome $\omega$, all real roots are simple for all sufficiently large $n$, so every asymptotic statement below remains valid if $N_n(I)$ is replaced by the multiplicity-counting quantity $R_n(I)$.

5.1. Obstruction: one-point concentration is not enough for the full sequence

Lemma 5.1. Let $c>0$. Then \[ \sum_{n=3}^\infty e^{-c\sqrt{\log n}}=\infty. \] Consequently, the presently available one-point concentration bound \[ \mathbb P\left(|N_n(I)-\mathbb E N_n(I)|\ge \varepsilon\log n\right) \le C_\varepsilon e^{-c_\varepsilon\sqrt{\log n}} \] cannot by itself yield the full-sequence almost sure convergence via Borel--Cantelli.

Proof. Consider the integral \[ \int_3^\infty e^{-c\sqrt{\log x}}\,dx. \] Make the substitution $u=\sqrt{\log x}$, so that $x=e^{u^2}$ and $dx=2u e^{u^2}\,du$. Then \[ \int_3^\infty e^{-c\sqrt{\log x}}\,dx =\int_{\sqrt{\log 3}}^\infty 2u\,e^{u^2-cu}\,du. \] Since $u^2-cu\to\infty$ as $u\to\infty$, the integrand $2u e^{u^2-cu}$ tends to $\infty$ and in particular is eventually bounded below by $1$. Hence the integral diverges. By the integral test, \[ \sum_{n=3}^\infty e^{-c\sqrt{\log n}}=\infty. \] The final claim is immediate: the first Borel--Cantelli lemma requires summability of the probabilities along the full sequence, and Lemma 5.1 shows that the current one-point upper bound does not provide this. Therefore any proof of the original full-sequence almost sure law must use more than one-point concentration; specifically, one needs some maximal inequality or other mechanism controlling correlations between nearby degrees. $\square$

5.2. A generic summable-tail subsequence principle

The following lemma is the main new technical device of this round.

Lemma 5.2 (summable-tail subsequence principle). Let $(X_n)_{n\ge2}$ be random variables and $(a_n)_{n\ge2}$ deterministic real numbers. Let $(b_n)_{n\ge2}$ be positive deterministic numbers with $b_n\to\infty$. Assume that for every integer $m\ge1$ there is a sequence $(p_{n,m})_{n\ge2}$ such that \[ \mathbb P\left(|X_n-a_n|\ge m^{-1}b_n\right)\le p_{n,m} \qquad(n\ge2). \] Let $(n_k)_{k\ge1}$ be an increasing sequence of integers such that for every $m\ge1$, \[ \sum_{k=1}^\infty p_{n_k,m}<\infty. \] Then \[ \frac{X_{n_k}-a_{n_k}}{b_{n_k}}\to0 \qquad\text{almost surely as }k\to\infty. \]

Proof. Fix $m\ge1$. Define \[ E_{k,m}:=\left\{|X_{n_k}-a_{n_k}|\ge m^{-1}b_{n_k}\right\}. \] By hypothesis, \[ \sum_{k=1}^\infty \mathbb P(E_{k,m})\le \sum_{k=1}^\infty p_{n_k,m}<\infty. \] The first Borel--Cantelli lemma implies \[ \mathbb P(E_{k,m}\ \text{i.o.})=0. \] Therefore, for almost every outcome $\omega$, there exists $K_m(\omega)$ such that for all $k\ge K_m(\omega)$, \[ |X_{n_k}(\omega)-a_{n_k}|<m^{-1}b_{n_k}. \] Now intersect over all $m\in\mathbb N$. Since this is a countable intersection of probability-one events, it still has probability one. On that event, for every $m\ge1$ the ratio $|X_{n_k}-a_{n_k}|/b_{n_k}$ is eventually at most $1/m$. Hence the ratio tends to $0$. $\square$

5.3. The relevant concentration input

We now record the specific concentration estimate that will be plugged into Lemma 5.2.

Lemma 5.3. For each fixed $\varepsilon>0$ and each interval \[ I\in\{\mathbb R,[0,1],[1,\infty),[-1,0],(-\infty,-1]\}, \] there exist constants $c_{\varepsilon,I},C_{\varepsilon,I}>0$ such that for all $n\ge2$, \[ \mathbb P\left(|N_n(I)-\mathbb E N_n(I)|\ge \varepsilon\log n\right) \le C_{\varepsilon,I}e^{-c_{\varepsilon,I}\sqrt{\log n}}. \]

Justification. This is precisely the consequence of Can--Nguyen's Theorems 1.4 and 1.5 recorded by Do in the proof of his Lemma 2.3. $\square$

5.4. Expectation constants for the four fundamental intervals

We next determine the asymptotic expectations on the intervals $[0,1]$, $[-1,0]$, $[1,\infty)$, and $(-\infty,-1]$.

Lemma 5.4. As $n\to\infty$, \[ \mathbb E N_n([0,1]) =\mathbb E N_n([-1,0]) =\mathbb E N_n([1,\infty)) =\mathbb E N_n((-\infty,-1]) =\frac{1}{2\pi}\log n+o(\log n). \]

Proof. We proceed in three steps.

Step 1: positive and negative roots inside $[-1,1]$ have the same expectation. Consider the polynomial \[ g_n(x):=f_n(-x)=\sum_{k=0}^n \epsilon_k(-1)^k x^k. \] The coefficient sequence $((-1)^k\epsilon_k)_{0\le k\le n}$ has the same joint distribution as $(\epsilon_k)_{0\le k\le n}$, because multiplying each coordinate by the deterministic sign $(-1)^k$ preserves the i.i.d. Rademacher law. Hence $g_n$ has the same distribution as $f_n$.

A real number $x\in[0,1]$ is a root of $g_n$ if and only if $-x\in[-1,0]$ is a root of $f_n$, with the same multiplicity. Therefore the random variables $N_n([0,1])$ and $N_n([-1,0])$ have the same distribution, hence \[ \mathbb E N_n([0,1])=\mathbb E N_n([-1,0]). \]

Step 2: inside and outside expectations agree on each sign side. By the Round-1 reciprocal symmetry, \[ \mathbb E N_n(0,1)=\mathbb E N_n(1,\infty). \] Adding the same endpoint contribution $\mathbb P(f_n(1)=0)$ to both sides gives \[ \mathbb E N_n([0,1])=\mathbb E N_n([1,\infty)). \] Similarly, \[ \mathbb E N_n([-1,0])=\mathbb E N_n((-\infty,-1]). \]

Step 3: evaluate the common asymptotic. Do records the classical expectation asymptotic \[ \mathbb E N_n([-1,1])=\frac{1}{\pi}\log n+o(\log n). \] Since $0$ is never a root of $f_n$ (because $f_n(0)=\epsilon_0\in\{\pm1\}$), the intervals $[-1,0]$ and $[0,1]$ split $[-1,1]$ without overlap on roots, and therefore \[ \mathbb E N_n([-1,1])=\mathbb E N_n([-1,0])+\mathbb E N_n([0,1]). \] By Step 1 the two summands are equal, so each equals \[ \frac12\,\mathbb E N_n([-1,1]) =\frac{1}{2\pi}\log n+o(\log n). \] Then Step 2 transfers the same asymptotic to $[1,\infty)$ and $(-\infty,-1]$. $\square$

5.5. Dense-subsequence strong laws

We can now combine Lemma 5.2 and Lemma 5.3 with the expectation asymptotics above.

Theorem 5.5 (dense-subsequence strong law on the four fundamental intervals and on $\mathbb R$). Let $(n_k)_{k\ge1}$ be an increasing sequence of integers such that \[ \frac{\sqrt{\log n_k}}{\log k}\to\infty. \] Then the following almost sure convergences hold: \[ \frac{N_{n_k}(\mathbb R)}{\log n_k}\to\frac{2}{\pi}, \] and for each interval \[ I\in\{[0,1],[-1,0],[1,\infty),(-\infty,-1]\} \] one has \[ \frac{N_{n_k}(I)}{\log n_k}\to\frac{1}{2\pi} \qquad\text{almost surely.} \] Consequently, \[ \frac{N_{n_k}((-\infty,-1]\cup[1,\infty))}{\log n_k}\to\frac{1}{\pi} \qquad\text{almost surely.} \]

Proof. We first prove the statement for a fixed interval $I$ among \[ \mathbb R,[0,1],[-1,0],[1,\infty),(-\infty,-1]. \] For each integer $m\ge1$, apply Lemma 5.3 with $\varepsilon=1/m$. Then there exist constants $c_{m,I},C_{m,I}>0$ such that \[ \mathbb P\left(\left|N_n(I)-\mathbb E N_n(I)\right|\ge m^{-1}\log n\right) \le C_{m,I}e^{-c_{m,I}\sqrt{\log n}} \qquad(n\ge2). \] Because \[ \frac{\sqrt{\log n_k}}{\log k}\to\infty, \] for each fixed $m$ there exists $K_m$ such that for all $k\ge K_m$, \[ c_{m,I}\sqrt{\log n_k}\ge 3\log k. \] Therefore, for all large $k$, \[ C_{m,I}e^{-c_{m,I}\sqrt{\log n_k}} \le C_{m,I}k^{-3}. \] Hence \[ \sum_{k=1}^\infty \mathbb P\left(\left|N_{n_k}(I)-\mathbb E N_{n_k}(I)\right|\ge m^{-1}\log n_k\right)<\infty. \] Lemma 5.2 now yields \[ \frac{N_{n_k}(I)-\mathbb E N_{n_k}(I)}{\log n_k}\to0 \qquad\text{almost surely.} \]

It remains to identify the deterministic limit of $\mathbb E N_{n_k}(I)/\log n_k$.

• For $I=\mathbb R$, Round 1 already used the expectation asymptotic of Do--Nguyen--Vu: \[ \mathbb E N_n(\mathbb R)=\frac{2}{\pi}\log n + C + o(1). \] Hence \[ \frac{\mathbb E N_{n_k}(\mathbb R)}{\log n_k}\to \frac{2}{\pi}. \] Combining with the previous display gives \[ \frac{N_{n_k}(\mathbb R)}{\log n_k}\to\frac{2}{\pi} \qquad\text{almost surely.} \]

• For $I\in\{[0,1],[-1,0],[1,\infty),(-\infty,-1]\}$, Lemma 5.4 gives \[ \frac{\mathbb E N_{n_k}(I)}{\log n_k}\to \frac{1}{2\pi}. \] Therefore \[ \frac{N_{n_k}(I)}{\log n_k}\to\frac{1}{2\pi} \qquad\text{almost surely.} \]

Finally, the intervals $[1,\infty)$ and $(-\infty,-1]$ are disjoint, so their counts add exactly. Hence \[ N_{n_k}((-\infty,-1]\cup[1,\infty)) = N_{n_k}((-\infty,-1]) + N_{n_k}([1,\infty)), \] and dividing by $\log n_k$ yields the outside-root limit $1/\pi$. $\square$

Corollary 5.6 (explicit non-lacunary subsequence with ratio tending to $1$). Define, for $k\ge3$, \[ \nu_k:=\exp\left((\log k)^3\right) \qquad\text{and}\qquad n_k:=\lfloor \nu_k\rfloor. \] After discarding finitely many initial terms, $(n_k)$ is strictly increasing and satisfies \[ \frac{n_{k+1}}{n_k}\to1. \] Hence Theorem 5.5 applies to this explicit non-lacunary subsequence. In particular, \[ \frac{N_{n_k}(\mathbb R)}{\log n_k}\to\frac{2}{\pi} \qquad\text{and}\qquad \frac{N_{n_k}((-\infty,-1]\cup[1,\infty))}{\log n_k}\to\frac{1}{\pi} \] almost surely.

Proof. We first verify the growth condition in Theorem 5.5. Since \[ \log n_k=(\log k)^3+o(1), \] we obtain \[ \frac{\sqrt{\log n_k}}{\log k} \sim \frac{(\log k)^{3/2}}{\log k} =(\log k)^{1/2}\to\infty. \] So Theorem 5.5 applies once $(n_k)$ is strictly increasing.

It remains to show $n_{k+1}/n_k\to1$. Because \[ \log \nu_k=(\log k)^3, \] we have \[ \log \frac{\nu_{k+1}}{\nu_k} =(\log(k+1))^3-(\log k)^3. \] By the mean value theorem applied to $\phi(x)=(\log x)^3$, there exists $\xi_k\in(k,k+1)$ such that \[ (\log(k+1))^3-(\log k)^3 =\phi'(\xi_k) =\frac{3(\log \xi_k)^2}{\xi_k}\to0. \] Hence $\nu_{k+1}/\nu_k\to 1$. Since $n_k/\nu_k\to1$, the same limit holds for $n_{k+1}/n_k$. Also, \[ \nu_{k+1}-\nu_k\sim \nu_k\cdot \frac{3(\log k)^2}{k}\to\infty, \] so after removing finitely many initial indices the floors $n_k$ are strictly increasing. $\square$

Corollary 5.7 (full-sequence liminf/limsup constraints). Almost surely, \[ \liminf_{n\to\infty}\frac{R_n}{\log n}\ge \frac{1}{\pi} \qquad\text{and}\qquad \limsup_{n\to\infty}\frac{R_n}{\log n}\ge \frac{2}{\pi}. \]

Proof. Since $R_n\ge N_n([-1,1])$ for every $n$, Do's strong law on $[-1,1]$ gives \[ \liminf_{n\to\infty}\frac{R_n}{\log n} \ge \lim_{n\to\infty}\frac{N_n([-1,1])}{\log n} =\frac{1}{\pi} \qquad\text{almost surely.} \] For the limsup, apply Theorem 5.5 to the explicit sequence from Corollary 5.6 (or alternatively use the Round-1 lacunary theorem): along that subsequence, \[ \frac{R_{n_k}}{\log n_k}\to \frac{2}{\pi} \qquad\text{almost surely,} \] so the full-sequence limsup is at least $2/\pi$ almost surely. $\square$

ADVERSARIAL VERIFICATION

I now check the new arguments against the main failure modes.

• Did I really go beyond Round 1? Yes. Round 1 proved only a lacunary almost sure law on $\mathbb R$. Theorem 5.5 replaces this by a much denser subsequence theorem, valid whenever $\sqrt{\log n_k}/\log k\to\infty$, and gives exact outside-half-line laws that were absent in Round 1.

• Was lacunarity accidentally used anywhere? No. The proof of Theorem 5.5 uses only the summability of \[ e^{-c_{m,I}\sqrt{\log n_k}}, \] which follows from $\sqrt{\log n_k}/\log k\to\infty$. No ratio lower bound $n_{k+1}/n_k\ge q>1$ is used.

• Is the condition $\sqrt{\log n_k}/\log k\to\infty$ strong enough for all rational error tolerances? Yes. For each fixed $m$, the constant $c_{m,I}>0$ is fixed. Hence eventually \[ c_{m,I}\sqrt{\log n_k}\ge 3\log k, \] which makes the $m$th error event summable in $k$. Countably many $m$ are handled by a countable intersection of probability-one events.

• Did endpoint roots at $\pm1$ create double-counting issues? They do not invalidate any asymptotic. In Lemma 5.4, the equality \[ \mathbb E N_n([0,1])=\mathbb E N_n([1,\infty)) \] is exact because both closed intervals differ from their open counterparts by the same endpoint event $\{f_n(1)=0\}$. Likewise on the negative side.

• Did multiplicities matter? The concentration theorems and expectation asymptotics are stated for distinct roots. By the Round-1 multiple-root lemma, real multiple roots occur only finitely often almost surely. Therefore every asymptotic statement in Theorem 5.5 and its corollaries remains valid for the original multiplicity-counting quantities $R_n(I)$.

• Any hidden use of unproved expectation asymptotics on $[1,\infty)$ or $(-\infty,-1]$? No. These were derived in Lemma 5.4 from (i) the classical expectation on $[-1,1]$, (ii) the sign symmetry $x\mapsto -x$, and (iii) the Round-1 reciprocal expectation symmetry.

• Does Corollary 5.6 really give a non-lacunary sequence? Yes. The ratio $n_{k+1}/n_k\to1$, so there is no fixed $q>1$ with $n_{k+1}\ge q n_k$ for all large $k$. Hence the sequence is genuinely non-lacunary.

No contradiction or gap was found in the new arguments.

FINAL

UNRESOLVED (BUT STRICTLY ADVANCED)

The original full-sequence statement remains unresolved. However, this round proves the following new theorems beyond Round 1.

1. Rigorous obstruction. The current one-point concentration estimate is non-summable along the full sequence, so Borel--Cantelli alone cannot prove the original almost sure law; a maximal inequality or other inter-degree control is genuinely necessary.

2. Dense-subsequence global law. If $(n_k)$ is any deterministic increasing sequence with \[ \frac{\sqrt{\log n_k}}{\log k}\to\infty, \] then \[ \frac{R_{n_k}}{\log n_k}\to\frac{2}{\pi} \qquad\text{almost surely.} \] This is strictly stronger than the Round-1 lacunary theorem.

3. Dense-subsequence law on the outside half-lines. Along the same class of subsequences, \[ \frac{R_{n_k}([1,\infty))}{\log n_k}\to\frac{1}{2\pi}, \qquad \frac{R_{n_k}((-\infty,-1])}{\log n_k}\to\frac{1}{2\pi}, \] and therefore \[ \frac{R_{n_k}((-\infty,-1]\cup[1,\infty))}{\log n_k}\to\frac{1}{\pi} \qquad\text{almost surely.} \]

4. Explicit non-lacunary subsequence. The same conclusions hold along the explicit deterministic sequence \[ n_k=\lfloor\exp\left((\log k)^3\right)\rfloor \] (after removing finitely many initial terms), for which $n_{k+1}/n_k\to1$.

Thus the remaining gap is now sharply localized: the unresolved issue is not one-point concentration, but the absence of a method that propagates these dense-subsequence laws across the residual windows between nearby degrees.

COMPLETION ESTIMATE

COMPLETION: 74%

REFERENCES

1. Y. Q. Do, A strong law of large numbers for real roots of random polynomials, arXiv:2403.06353.
2. V. H. Can and O. Nguyen, Concentration inequalities for the number of real zeros of Kac polynomials, arXiv:2311.15446.
3. Y. Q. Do, H. Nguyen, and V. Vu, Real roots of random polynomials: expectation and repulsion, arXiv:1409.4128.
4. T. F. Bloom, Erdős Problem #521, erdosproblems.com/521 (checked for current open status).

Erdős Problem #521: independent continuation

1) TARGET AND SOURCE AUDIT

All supplied versions considered: 521.tex, 521_v2.tex. Version 1 proves a deterministic subsequence consequence of one-point concentration. Version 2 proves convergence outside a random set of finite reciprocal sum, but labels the result only as logarithmic-density-one. In fact it implies ordinary density-one convergence; I prove that upgrade and its limit.

2) WORK

A deterministic strengthening.

For completeness, the supplied concentration hypothesis $\mathbb P(|R_n/\log n-2/\pi|>\epsilon)\le C_\epsilon e^{-c_\epsilon\sqrt{\log n}}$ implies finite exceptional reciprocal sum by Tonelli, since substitution $u=\sqrt{\log x}$ gives $\int_3^\infty e^{-c\sqrt{\log x}}dx/x=\int_{\sqrt{\log3}}^\infty2ue^{-cu}du<\infty$. A diagonal choice of cutoffs for tolerances $1/j$ gives a single $G$: choose each tail reciprocal mass below $2^{-j}$, omit the corresponding bad points in successive cutoff blocks, and also allow the finitely many initial points. The initial finite set was omitted from one displayed estimate in version 2, but it has no effect on finiteness.

This still cannot give convergence at every degree: a deterministic sequence equal to $c+1$ at $n=2^j$ and $c$ otherwise has a harmonic-summable exceptional set and fails to converge.

3) ADVERSARIAL VERIFICATION

4) FINAL

5) SOURCES CHECKED

6) COMPLETION ESTIMATE