An Explicit Window for Hypothetical Colossally Abundant Counterexamples to Robin's Criterion

1. Introduction

The Riemann hypothesis (RH), concerning the nontrivial zeros of the Riemann zeta function, has been called the “Holy Grail of Mathematics” [1,2]. Numerous equivalent formulations exist [3]; one of particular interest here is Robin’s criterion [4], which asserts that RH is equivalent to the inequality

$$\sigma \left(n\right)<e^{\gamma }·n·loglogn$$

holding for all $n>5040$. Here, $\gamma \approx 0.57721$ is the Euler–Mascheroni constant, $\sigma$ is the divisor sum function, log denotes the natural logarithm, and n is a natural number.

Among the integers, the superabundant and colossally abundant numbers, introduced by Ramanujan [5] and studied by Alaoglu and Erdos [6], play a central role in the analysis of Robin’s inequality. By a theorem of Robin [[4] Proposition 1, p. 204], if RH is false then there exist infinitely many colossally abundant numbers $n>5040$ for which Robin’s inequality fails. Since every colossally abundant number is superabundant, ruling out colossally abundant counterexamples for $n>5040$ is therefore equivalent to proving RH.

In this work we make no such claim. Rather, we use the explicit prime-counting bounds of Aoudjit, Berkane, and Dusart [7] together with structural properties of colossally abundant numbers to derive an unconditional, effective upper bound on the size of any hypothetical colossally abundant counterexample (Theorem 1). The bound takes the form $n<N_k^{Y_k}$, where $N_k$ is the primorial of order k associated to the largest prime factor $p_k$ of n, and $Y_k$ is an explicit constant satisfying $Y_k\to 1^{+}$ as $k\to \infty$. Combined with the trivial lower bound $n\ge N_k$ for Hardy–Ramanujan integers, this confines any hypothetical colossally abundant counterexample to a narrow explicit interval above its primorial. We then deduce a conditional reformulation of RH localized to this interval (Theorem 2).

By restricting attention to colossally abundant numbers we obtain access to the explicit exponent formula of Alaoglu–Erdos [[6] Theorem 10, p. 462], which would be unavailable in the broader superabundant setting. Even with this additional structure, however, the manuscript does not establish the Riemann hypothesis. The combination of this exponent formula with Axler’s two-sided Chebyshev bound (Proposition 5) controls the smooth part of $logn$ at the precision required by Theorem 1, but a residual fractional-part sum—an irregular contribution that no Chebyshev-type estimate can address—remains uncontrolled at the relevant scale. We discuss this obstruction in Section 3.6.

2. Background and Ancillary Results

The Euler–Mascheroni constant, denoted $\gamma \approx 0.57721$, is defined as

$$\gamma =\underset{n\to \infty }{lim}\left(H_n-logn\right),$$

where log denotes the natural logarithm and $H_n={\sum }_{k=1}^n{\displaystyle \frac{1}{k}}$ is the $n^{\mathrm{th}}$ harmonic number. As usual, $\sigma \left(n\right)$ denotes the sum of all positive divisors of n:

$$\sigma \left(n\right)=\sum _{d\mid n}d,$$

where $d\mid n$ means that the integer d divides n. We define the abundancy index$I:\mathbb{N}\to \mathbb{Q}$ by $I\left(n\right)={\displaystyle \frac{\sigma \left(n\right)}{n}}$. Since $\sigma$ is multiplicative, $I\left(n\right)$ admits a product representation:

Proposition 1. 

Let $n={\prod }_{i=1}^r{p}_i^{a_i}$ be the prime factorization of n, where $p_1<\dots <p_r$ are distinct primes and $a_1,\dots ,a_r$ are positive integers. Then [[8]Lemma 1 (2) pp. 2]:

$$\begin{array}{cc}\hfill I\left(n\right)& =\left({\displaystyle \prod _{i=1}^r}{\displaystyle \frac{p_i}{p_i-1}}\right)·{\displaystyle \prod _{i=1}^r}\left(1-{\displaystyle \frac{1}{p_i^{a_i+1}}}\right)\hfill \\ \hfill & ={\displaystyle \prod _{i=1}^r}I\left(p_i^{a_i}\right).\hfill \end{array}$$

Proposition 2. 

For $n>1$ [[9] (2.7) pp. 362]:

$$I\left(n\right)<\prod _{p\mid n}{\displaystyle \frac{p}{p-1}}.$$

We introduce the following shorthand for Robin’s inequality:

Definition 1. 

We say that $\mathsf{Robin}\left(n\right)$ holds if

$$I\left(n\right)<e^{\gamma }·loglogn.$$

The Chebyshev function $\theta \left(x\right)$ is defined by

$$\theta \left(x\right)=\sum _{p\le x}logp,$$

where the sum is taken over all primes $p\le x$. The following two propositions provide the quantitative bounds on $\theta$ and on the Mertens product ${\prod }_{p\le x}{\displaystyle \frac{p}{p-1}}$ that we will use:

Proposition 3. 

For $x\ge 7,232,121,212$ [[7] Lemma 2.7 (4) pp. 19]:

$$\theta \left(x\right)\ge \left(1-{\displaystyle \frac{0.01}{{log}^3\left(x\right)}}\right)·x.$$

Proposition 4. 

For $x\ge 2,278,382$ [[7] Lemma 2.7 (5) pp. 19]:

$$\prod _{p\le x}{\displaystyle \frac{p}{p-1}}\le e^{\gamma }·(logx)·\left(1+{\displaystyle \frac{0.2}{{log}^3\left(x\right)}}\right).$$

A two-sided refinement of the Chebyshev estimate, due to Axler, will be used in the discussion of the obstruction in Section 3.6. It improves on Proposition 3 both by being two-sided and by enlarging the constant in the error term to $0.15$.

Proposition 5. 

For every $x\ge 19,035,709,163$ [[10] Theorem 1 pp. 2]:

$$\left(1-{\displaystyle \frac{0.15}{{log}^{3}x}}\right)·x<\theta \left(x\right)<\left(1+{\displaystyle \frac{0.15}{{log}^{3}x}}\right)·x.$$

Ramanujan’s theorem asserts that RH implies $\mathsf{Robin}\left(n\right)$ for all sufficiently large n [5]. Robin’s theorem sharpens this to a precise equivalence:

Proposition 6. 

$\mathsf{Robin}\left(n\right)$ holds for all natural numbers $n>5040$ if and only if the Riemann hypothesis is true [[4] Theorem 1 pp. 188].

Ramanujan’s unpublished notes [5] introduced generalized highly composite numbers, encompassing both superabundant and colossally abundant numbers; these were also studied by Alaoglu and Erdos [6]. Given the first k consecutive primes $p_1=2,p_2=3,\dots ,p_k$, an integer of the form ${\prod }_{i=1}^k{p}_i^{a_i}$ with $a_1\ge a_2\ge \dots \ge a_k\ge 1$ is called a Hardy–Ramanujan integer [[9] pp. 367]. A natural number n is called superabundant if $I\left(m\right)<I\left(n\right)$ for every natural number $m<n$. A natural number n is called colossally abundant if there exists $ϵ>0$ such that

$${\displaystyle \frac{\sigma \left(n\right)}{n^{1+ϵ}}}\ge {\displaystyle \frac{\sigma \left(m\right)}{m^{1+ϵ}}}\mathrm{for}\mathrm{all}m\ge 1.$$

Proposition 7. 

Every colossally abundant number is superabundant [[6] pp. 455], and every superabundant number is a Hardy–Ramanujan integer [[6] Theorem 1 pp. 450]. In particular, every colossally abundant number is a Hardy–Ramanujan integer.

Proposition 8. 

If n is superabundant with largest prime factor p, then [[6] Theorem 7 pp. 454]

$$p\sim logn(n\to \infty ).$$

In particular, the largest prime factor of a colossally abundant number grows without bound along any infinite sequence of distinct colossally abundant numbers.

The next two propositions describe the structure of hypothetical counterexamples:

Proposition 9. 

If $n>5040$ is the smallest integer for which $\mathsf{Robin}\left(n\right)$ fails, then the largest prime factor p of n satisfies $p>e^{31.018189471}$ [[11] Theorem 4.2 pp. 748].

Proposition 10. 

If the Riemann hypothesis is false, then there exist infinitely many colossally abundant numbers $n>5040$ for which $\mathsf{Robin}\left(n\right)$ fails [[4] Proposition 1 pp. 204].

The following explicit formula of Alaoglu and Erdos describes the prime-power exponents of a colossally abundant number in terms of its parameter $ϵ$. We will not use it in the proof of our main bound, but it will be relevant to the discussion of the obstruction in Section 3.6.

Proposition 11. 

Let $n_{ϵ}$ be the colossally abundant number associated with parameter $ϵ>0$, and let $k_q\left(ϵ\right)$ denote the exponent of the prime q in the prime factorization of $n_{ϵ}$. Then [[6] Theorem 10 pp. 462]:

$$k_q\left(ϵ\right)=⌊{\displaystyle \frac{log\left(\frac{q^{1+ϵ}-1}{q^{ϵ}-1}\right)}{logq}}⌋-1.$$

3. Main Results

Definition 2. 

Thep-adic order of a nonzero integer n, denoted ${\nu }_p\left(n\right)$, is the exponent of the highest power of the prime p that divides n.

3.1. A Structural Lemma for Colossally Abundant Numbers

The following lemma sharpens, for colossally abundant n, the relationship between the largest prime factor of n and the prime power $q^{{\nu }_q\left(n\right)+1}$ for any prime $q\le p$. The proof carries over verbatim from the corresponding statement for superabundant numbers, since the comparison argument uses only superabundance, and every colossally abundant number is superabundant by Proposition 7.

Lemma 1. 

Let n be a colossally abundant number with largest prime factor p. Then

$$p\le q^{{\nu }_q\left(n\right)+1}$$

for every prime q with $2\le q\le p$.

Proof. 

Suppose for contradiction that $p>q^{{\nu }_q\left(n\right)+1}$ for some prime $q\le p$. Set $m={\displaystyle \frac{n·q^{{\nu }_q\left(n\right)+1}}{p}}$. Since $m<n$ and n is superabundant by Proposition 7, we have $I\left(m\right)<I\left(n\right)$, i.e. $I\left(n\right)/I\left(m\right)>1$. Since I is multiplicative (Proposition 1), the ratio $I\left(n\right)/I\left(m\right)$ depends only on the prime powers at which n and m differ, namely q and p. Computing via Proposition 1:

$$\begin{array}{cc}\hfill 1& <{\displaystyle \frac{I\left(n\right)}{I\left(m\right)}}\hfill \\ \hfill & =\left({\displaystyle \frac{q^{2{\nu }_q\left(n\right)+2}-q^{{\nu }_q\left(n\right)+1}}{q^{2{\nu }_q\left(n\right)+2}-1}}\right)·\left(1+{\displaystyle \frac{1}{p}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{1+\frac{1}{q^{{\nu }_q\left(n\right)+1}}}}·\left(1+{\displaystyle \frac{1}{p}}\right),\hfill \end{array}$$

where the last equality follows from the factorization:

$$\begin{array}{cc}\hfill {\displaystyle \frac{q^{2{\nu }_q\left(n\right)+2}-q^{{\nu }_q\left(n\right)+1}}{q^{2{\nu }_q\left(n\right)+2}-1}}& =1-{\displaystyle \frac{q^{{\nu }_q\left(n\right)+1}-1}{q^{2{\nu }_q\left(n\right)+2}-1}}\hfill \\ \hfill & =1-{\displaystyle \frac{q^{{\nu }_q\left(n\right)+1}-1}{(q^{{\nu }_q\left(n\right)+1}-1)(q^{{\nu }_q\left(n\right)+1}+1)}}\hfill \\ \hfill & =1-{\displaystyle \frac{1}{q^{{\nu }_q\left(n\right)+1}+1}}\hfill \\ \hfill & ={\displaystyle \frac{q^{{\nu }_q\left(n\right)+1}}{q^{{\nu }_q\left(n\right)+1}+1}}\hfill \\ \hfill & ={\displaystyle \frac{1}{1+\frac{1}{q^{{\nu }_q\left(n\right)+1}}}}.\hfill \end{array}$$

Therefore ${\displaystyle \frac{1}{q^{{\nu }_q\left(n\right)+1}}}<{\displaystyle \frac{1}{p}}$, which gives $p<q^{{\nu }_q\left(n\right)+1}$, contradicting our assumption. □

Lemma 2. 

Let p be a fixed prime, and let n range over an infinite sequence of distinct colossally abundant numbers. Then ${\nu }_p\left(n\right)\to \infty$ as $n\to \infty$ along this sequence.

Proof. 

By Proposition 8, the largest prime factor P of n satisfies $P\to \infty$. For any fixed prime p, all sufficiently large terms of the sequence have $P\ge p$. Applying Lemma 1 with $q=p$ then gives

$$P\le p^{{\nu }_p\left(n\right)+1},\mathrm{i}.\mathrm{e}.,{\nu }_p\left(n\right)\ge {\displaystyle \frac{logP}{logp}}-1.$$

Since $P\to \infty$, the right-hand side grows without bound, so ${\nu }_p\left(n\right)\to \infty$. □

3.2. An Explicit Constant

Definition 3. 

For each prime $p_k>2$, define

$$Y_k={\displaystyle \frac{e^{\frac{0.2}{{log}^2\left(p_k\right)}}}{1-\frac{0.01}{{log}^3\left(p_k\right)}}}.$$

Lemma 3. 

The sequence $\left(Y_k\right)$ is strictly decreasing on the range $p_k>e$ and satisfies $Y_k\to 1^{+}$ as $k\to \infty$. In particular, $1<Y_k<1.00021$ for all $p_k>e^{31.018189471}$.

Proof. 

As $p_k$ increases, both ${\displaystyle \frac{0.2}{{log}^2\left(p_k\right)}}$ and ${\displaystyle \frac{0.01}{{log}^3\left(p_k\right)}}$ decrease monotonically to zero. Hence the numerator $e^{0.2/{log}^2\left(p_k\right)}$ decreases toward 1 from above, and the denominator $1-0.01/{log}^3\left(p_k\right)$ increases toward 1 from below; both are positive on the stated range, so $Y_k$ is strictly decreasing with limit 1. Strictness $Y_k>1$ is immediate: $e^{0.2/{log}^2\left(p_k\right)}>1$ while $0<1-0.01/{log}^3\left(p_k\right)<1$, so the ratio exceeds 1. The numerical bound $Y_k<1.00021$ for $p_k>e^{31.018189471}$ follows by direct evaluation at the threshold combined with monotonicity. □

3.3. The Key Prime-Product Estimate

The following lemma combines the bounds of Propositions 3 and 4 into a single inequality whose right-hand side is conveniently expressed in terms of $Y_k$ and $\theta \left(p_k\right)$.

Lemma 4. 

Let $p_1<p_2<\dots <p_k$ be the first k consecutive primes with $p_k>7,232,121,212$. Then

$${\displaystyle \prod _{i=1}^k}{\displaystyle \frac{p_i}{p_i-1}}\le e^{\gamma }·log\left(Y_k·\theta \left(p_k\right)\right).$$

Proof. 

By Proposition 3:

$$\theta \left(p_k\right)\ge \left(1-{\displaystyle \frac{0.01}{{log}^3\left(p_k\right)}}\right)·p_k.$$

Therefore:

$$\begin{array}{cc}\hfill log\left(Y_k·\theta \left(p_k\right)\right)& \ge log\left(Y_k·\left(1-{\displaystyle \frac{0.01}{{log}^3\left(p_k\right)}}\right)·p_k\right)\hfill \\ \hfill & =log{p}_k+log\left(Y_k·\left(1-{\displaystyle \frac{0.01}{{log}^3\left(p_k\right)}}\right)\right).\hfill \end{array}$$

Substituting the definition of $Y_k$ and canceling:

$$\begin{array}{cc}\hfill log\left(Y_k·\left(1-{\displaystyle \frac{0.01}{{log}^3\left(p_k\right)}}\right)\right)& =log\left({\displaystyle \frac{e^{\frac{0.2}{{log}^2\left(p_k\right)}}}{\left(1-\frac{0.01}{{log}^3\left(p_k\right)}\right)}}·\left(1-{\displaystyle \frac{0.01}{{log}^3\left(p_k\right)}}\right)\right)\hfill \\ \hfill & =log\left(e^{{\displaystyle \frac{0.2}{{log}^2\left(p_k\right)}}}\right)\hfill \\ \hfill & ={\displaystyle \frac{0.2}{{log}^2\left(p_k\right)}}.\hfill \end{array}$$

Combining:

$$log\left(Y_k·\theta \left(p_k\right)\right)\ge log{p}_k+{\displaystyle \frac{0.2}{{log}^2\left(p_k\right)}}.$$

Applying Proposition 4, and using $log{p}_k/{log}^3\left(p_k\right)=1/{log}^2\left(p_k\right)$ to expand the bound, then yields:

$${\displaystyle \prod _{i=1}^k}{\displaystyle \frac{p_i}{p_i-1}}\le e^{\gamma }·log{p}_k·\left(1+{\displaystyle \frac{0.2}{{log}^3\left(p_k\right)}}\right)=e^{\gamma }·\left(log{p}_k+{\displaystyle \frac{0.2}{{log}^2\left(p_k\right)}}\right)\le e^{\gamma }·log\left(Y_k·\theta \left(p_k\right)\right).$$

□

3.4. An Effective Upper Bound on Counterexamples

We now establish the central unconditional estimate of this paper.

Theorem 1. 

Let $n>5040$ be a colossally abundant number for which $\mathsf{Robin}\left(n\right)$ fails, and let $p_k$ be the largest prime factor of n. Assume $p_k>7,232,121,212$. Set $N_k={\prod }_{i=1}^k{p}_i$, the primorial of order k. Then

$$N_k\le n<{\left(N_k\right)}^{Y_k}.$$

Proof. 

By Proposition 7, n is a Hardy–Ramanujan integer, so $n={\prod }_{i=1}^k{p}_i^{a_i}$ with $a_1\ge \dots \ge a_k\ge 1$. The lower bound $N_k\le n$ follows immediately from $a_i\ge 1$ for every i. For the upper bound, the hypothesis $p_k>7,232,121,212$ ensures that Lemma 4 applies.

Since $\mathsf{Robin}\left(n\right)$ fails:

$$I\left(n\right)\ge e^{\gamma }·loglogn.$$

(1)

Noting that $log{\left(N_k\right)}^{Y_k}=Y_k·\theta \left(p_k\right)$, Lemma 4 gives:

$$\prod _{p\le p_k}{\displaystyle \frac{p}{p-1}}\le e^{\gamma }·loglog\left({\left(N_k\right)}^{Y_k}\right).$$

(2)

Chaining Proposition 2 (using that the prime divisors of n are exactly $p_1,\dots ,p_k$) with (1) and (2):

$$\begin{array}{cc}\hfill e^{\gamma }·loglog\left({\left(N_k\right)}^{Y_k}\right)& \ge \prod _{p\le p_k}{\displaystyle \frac{p}{p-1}}\hfill \\ \hfill & >I\left(n\right)\hfill \\ \hfill & \ge e^{\gamma }·loglogn.\hfill \end{array}$$

Since $x\mapsto loglogx$ is strictly increasing on $x>1$, we conclude ${\left(N_k\right)}^{Y_k}>n$. □

Corollary 1. 

If $n>5040$ is a colossally abundant counterexample to $\mathsf{Robin}$ with largest prime factor $p_k>7,232,121,212$, then

$$\theta \left(p_k\right)\le logn<Y_k·\theta \left(p_k\right).$$

The width of this window satisfies $logn-\theta \left(p_k\right)<(Y_k-1)\theta \left(p_k\right)$, where $Y_k-1<0.00021$ for all $p_k>e^{31.018189471}$.

Proof. 

Take logarithms in Theorem 1 and use $log{N}_k=\theta \left(p_k\right)$. The numerical estimate is Lemma 3. □

3.5. A Conditional Reformulation of the Riemann Hypothesis

Theorem 1 confines every hypothetical colossally abundant counterexample to a narrow explicit interval above its primorial. Combining this with Robin’s theorem yields the following equivalent reformulation.

Theorem 2. 

The following are equivalent:

(i)

The Riemann hypothesis is true.

(ii)

Every colossally abundant number n whose largest prime factor $p_k$ satisfies $p_k>7,232,121,212$ and which lies in the interval $N_k\le n<{\left(N_k\right)}^{Y_k}$ also satisfies $\mathsf{Robin}\left(n\right)$.

Proof. 

(i) ⇒ (ii). Assume RH. By Proposition 6, $\mathsf{Robin}\left(n\right)$ holds for every $n>5040$. Any n as in (ii) satisfies $n\ge N_k\ge p_k>7,232,121,212>5040$, so $\mathsf{Robin}\left(n\right)$ holds.

(ii) ⇒ (i). We argue by contrapositive. Assume RH is false. By Proposition 10, there exist infinitely many colossally abundant numbers $n>5040$ for which $\mathsf{Robin}\left(n\right)$ fails. By Proposition 8, the largest prime factor of a colossally abundant number tends to infinity along any infinite sequence of distinct colossally abundant numbers; hence among these infinitely many counterexamples we may choose one, call it $n_0$, whose largest prime factor $p_{k_0}$ satisfies $p_{k_0}>7,232,121,212$. By Theorem 1 applied to $n_0$,

$$N_{k_0}\le n_0<{\left(N_{k_0}\right)}^{Y_{k_0}}.$$

Thus $n_0$ is a colossally abundant number satisfying every hypothesis of (ii) yet failing $\mathsf{Robin}\left(n_0\right)$, contradicting (ii). □

Corollary 2. 

The Riemann hypothesis is equivalent to the assertion that for every k with $p_k>7,232,121,212$ and every colossally abundant Hardy–Ramanujan integer $n={\prod }_{i=1}^k{p}_i^{a_i}$ with $a_1\ge \dots \ge a_k\ge 1$ satisfying

$$\theta \left(p_k\right)\le logn<Y_k·\theta \left(p_k\right),$$

the inequality $I\left(n\right)<e^{\gamma }loglogn$ holds.

Proof. 

Combine Theorem 2 with Proposition 7 and Corollary 1. Taking logarithms of $N_k\le n<{\left(N_k\right)}^{Y_k}$ recasts the window in terms of $\theta \left(p_k\right)$, since $log{N}_k=\theta \left(p_k\right)$. □

3.6. The Remaining Obstruction

For completeness, we record the asymptotic obstruction that prevents Theorem 2 from yielding an unconditional proof of RH along the lines pursued here. We assume throughout this subsection that $p_k\ge 19,035,709,163$, so that Proposition 5 applies; this threshold is consistent with $p_k>e^{31.018189471}$ since $e^{31.018189471}>19,035,709,163$.

If $n={\prod }_{i=1}^k{p}_i^{a_i}$ is a colossally abundant counterexample, the structural inequality of Lemma 1 (with $q=p_i$) gives $a_i\ge \frac{log{p}_k}{log{p}_i}-1$, hence

$$logn={\displaystyle \sum _{i=1}^k}{a}_{i}log{p}_i\ge klog{p}_k-\theta \left(p_k\right).$$

By Abel summation, $klog{p}_k-\theta \left(p_k\right)={\int }_2^{p_k}\frac{\pi \left(t\right)}{t}dt$, and the prime number theorem gives

$$klog{p}_k-\theta \left(p_k\right)\sim {\displaystyle \frac{p_k}{log{p}_k}}(k\to \infty ).$$

On the other hand, $(Y_k-1)\theta \left(p_k\right)\sim 0.2p_k/{log}^2\left(p_k\right)$. Both quantities are $o\left(\theta \left(p_k\right)\right)$, so the structural lower bound $logn\ge klog{p}_k-\theta \left(p_k\right)$ does not contradict the upper bound $logn<Y_k·\theta \left(p_k\right)$ provided by Theorem 1.

The colossally abundant setting permits a refinement: by Proposition 11, the exponent of each prime $q\le p_k$ in the factorization of $n=n_{ϵ}$ is given exactly by

$$a_q=k_q\left(ϵ\right)=⌊x_q\left(ϵ\right)⌋-1,x_q\left(ϵ\right):={\displaystyle \frac{log\left(\frac{q^{1+ϵ}-1}{q^{ϵ}-1}\right)}{logq}},$$

where $ϵ$ is the parameter associated to $n_{ϵ}$. Writing $\left\{x\right\}=x-\lfloor x\rfloor$ for the fractional part,

$$logn=\underset{S\left(ϵ\right)}{\underbrace{\sum _{q\le p_k}{x}_q\left(ϵ\right)logq}}-\underset{F\left(ϵ\right)}{\underbrace{\sum _{q\le p_k}\left\{x_q\left(ϵ\right)\right\}logq}}-\theta \left(p_k\right).$$

(3)

Here $S\left(ϵ\right)$ is the smooth (continuous-in-$ϵ$) part and $-F\left(ϵ\right)-\theta \left(p_k\right)$ collects the floor and the explicit “$-1$” from Proposition 11. We have $0\le F\left(ϵ\right)<\theta \left(p_k\right)$.

3.6.1. The Smooth Part is Well-Controlled by Axler’s Bound

Since $a_{p_k}\ge 1$, the parameter $ϵ$ satisfies $ϵ\asymp 1/(p_{k}log{p}_k)$. Expanding $x_q\left(ϵ\right)$ in $ϵ$ and summing, one obtains

$$S\left(ϵ\right)=klog{p}_k+kloglog{p}_k+S_1\left(ϵ\right),$$

where $S_1\left(ϵ\right)$ is a smooth correction expressible as a linear combination of partial sums of $logq$ and 1 over primes $q\le p_k$. Such partial sums are precisely the quantities controlled by Proposition 5:

$$\theta \left(p_k\right)=p_k\left(1+O(1/{log}^3{p}_k)\right),\pi \left(p_k\right)={\displaystyle \frac{p_k}{log{p}_k}}\left(1+O(1/log{p}_k)\right),$$

with the latter following from the former by partial summation. In particular, $S_1\left(ϵ\right)$ admits an effective two-sided bound at scale $o(p_k/{log}^2{p}_k)$ from Proposition 5. Hence the smooth part of (3) can be evaluated to within the precision required by Theorem 1.

3.6.2. The Floor Contribution is the Obstruction

The remaining term $F\left(ϵ\right)={\sum }_{q\le p_k}\left\{x_q\left(ϵ\right)\right\}logq$ is not controlled by any Chebyshev-type estimate, because the fractional parts $\left\{x_q\left(ϵ\right)\right\}$ depend on the arithmetic of $ϵ$ relative to $logq$ for each prime q individually. The only universal bound is the trivial one,

$$0\le F\left(ϵ\right)<\theta \left(p_k\right)\sim p_k,$$

which is too weak by a factor of ${log}^2{p}_k$ to determine whether

$$logn-\theta \left(p_k\right)=\left(klog{p}_k-\theta \left(p_k\right)\right)+kloglog{p}_k+S_1\left(ϵ\right)-F\left(ϵ\right)$$

exceeds $(Y_k-1)\theta \left(p_k\right)\sim 0.2p_k/{log}^2{p}_k$. While the leading explicit positive terms

$$\left(klog{p}_k-\theta \left(p_k\right)\right)+kloglog{p}_k\sim {\displaystyle \frac{p_k}{log{p}_k}}+{\displaystyle \frac{p_{k}loglog{p}_k}{log{p}_k}}$$

exceed $0.2p_k/{log}^2{p}_k$ asymptotically, the term $-F\left(ϵ\right)$ may be of size up to $-\theta \left(p_k\right)\sim -p_k$, dwarfing both the leading positive contributions and the window width. The sign of $logn-\theta \left(p_k\right)$ relative to $(Y_k-1)\theta \left(p_k\right)$ therefore cannot be determined without an effective estimate for $F\left(ϵ\right)$.

3.6.3. Reformulating the Obstruction

The above analysis shows that the obstruction to closing the gap is not a deficiency of the Chebyshev estimate—which Axler’s bound (Proposition 5) handles to the required precision—but the absence of a sufficiently sharp upper bound on the irregular sum

$$F\left(ϵ\right)=\sum _{q\le p_k}\left\{x_q\left(ϵ\right)\right\}logq$$

for parameters $ϵ\asymp 1/(p_{k}log{p}_k)$ corresponding to colossally abundant counterexamples. Concretely, an unconditional refinement of Theorem 2 sufficient to establish RH along these lines would require a bound of the form

$$F\left(ϵ\right)\le \left(klog{p}_k-\theta \left(p_k\right)\right)+kloglog{p}_k-0.2p_k/{log}^2{p}_k+o(p_k/{log}^2{p}_k)$$

to fail for the relevant $ϵ$. Since the right-hand side is asymptotic to $p_{k}loglog{p}_k/log{p}_k$, this is not a numerically severe constraint; but $F\left(ϵ\right)$ depends on the joint distribution of the residues $\left\{x_q\left(ϵ\right)\right\}$ across primes $q\le p_k$, and no unconditional method is known to control such fractional-part sums at this scale uniformly in $ϵ$. Establishing the required estimate appears to require techniques outside the scope of explicit prime-counting bounds, and we leave it as an open problem.