bibkey: coffey2009stieltjes
authors: Mark W. Coffey
year: 2009
title: The Stieltjes constants, their relation to the eta_j coefficients, and representation of the Hurwitz zeta function
doi: null
url: https://arxiv.org/abs/0706.0343v2
claim: Proposition 2(a) splits the Li arithmetic contribution into a Stieltjes term and an ordered von Mangoldt sum; standard orthogonal coefficient budgets control the Stieltjes and proper-prime-power channels, while the centered primary-prime joint estimate remains unproved.
strata_touched: []
license: citation-only
triage: anchor

Stieltjes and prime-power parts of the Li joint estimate

The primary is Mark W. Coffey, arXiv:0706.0343v2, version 25 February 2009, first submitted in 2007. Source locators below refer to that version. Proposition 2(a), equations (12), (14)–(17), supplies the decomposition. The integral representation and orthogonal coefficient budgets below are applications of standard Euler–Maclaurin and Laguerre identities, not a new version of those results or a priority claim. The adjacent-Fibonacci consumer reuses Lagarias’s Weil kernel. This is paper mathematics without Lean certification; the actual signed prime estimate and RH remain unproved.

CS1. Exact normalization and the ordered prime-power term

Use the classical constants with conventions

$$\zeta (1+s)=\frac{1}{s}+\sum _{k\ge 0}\frac{(-1{)}^k{\gamma }_k}{k!}{s}^k,-\frac{{\zeta }^{\prime }}{\zeta }(1+s)=\frac{1}{s}+\sum _{k\ge 0}{\eta }_k{s}^k.$$

Coffey’s Proposition 2(a) gives, for each integer $n\ge 1$,

$$S_2(n)=-\sum _{r=1}^n\left(\genfrac{}{}{0ex}{}{n}{r}\right){\eta }_{r-1}=S_{\gamma }(n)+S_{\Lambda }(n),S_{\gamma }(n)=\sum _{r=1}^n\frac{(-1{)}^r}{(r-1)!}\left(\genfrac{}{}{0ex}{}{n}{r}\right){\gamma }_{r-1},\text{\hspace{1em}(CS1)}$$

$$S_{\Lambda }(n)=n+\underset{X\to \infty }{\lim }\sum _{2\le m\le X}\frac{1-\Lambda (m)}{m}{L}_{n-1}^{(1)}(\log m).\text{\hspace{1em}(CS2)}$$

Here $\Lambda (p^k)=\log p$ for every prime power, and vanishes otherwise. The sum is in increasing integer order, not an absolutely convergent prime-only sum. For fixed $n$, its convergence follows by partial summation from the existing unconditional PNT supplier, which in particular gives the classical weaker remainder ${\sum }_{m\le x}(\Lambda (m)-1)=O(x{e}^{-c\sqrt{\log x}})$ for some $c>0$. No uniform-in-$n$ estimate follows from this fixed-polynomial argument. The $m=1$ contribution is exactly $L_{n-1}^{(1)}(0)=n$ and is retained.

Let $H_n=S_{\infty }(n)+1$ in the classical specialization of Lagarias, published Lemma 4.4, equations (4.5)–(4.8). The exact split is ${\lambda }_n=H_n+S_2(n)$. Put

$$U_n=n+S_{\gamma }(n),P_n=\underset{X\to \infty }{\lim }\sum _{2\le m\le X}\frac{\Lambda (m)-1}{m}{L}_{n-1}^{(1)}(\log m).$$

Then, with the same Li normalization as the Fibonacci growth application,

$$\boxed{{\lambda }_n=H_n+U_n-P_n.}\text{\hspace{1em}(CS3)}$$

All three indices used below refer to this single actual arithmetic sequence. They are not independently optimized prime configurations.

CS2. Source-use scope of the reported linear bound

Proposition 2(b) reports $S_{\gamma }(n)=O(n)$. Its displayed proof, equation (52), replaces a bounded periodic Bernoulli factor by a constant and then integrates a signed Laguerre polynomial. That step does not bound the absolute integral: the polynomial changes sign. Equations (49) and (51) also print boundary signs that are not adopted here. These observations do not establish that Proposition 2(b) is false. Its linear-growth conclusion follows independently from the standard orthogonal argument below; the displayed source majorization is not used. The stronger Corollary 16 rate is not imported.

CS3. A finite orthogonal budget for the Stieltjes channel

Write $P_1(t)=\{t\}-1/2$. The standard Euler–Maclaurin formula DLMF 25.11.5, with $a=1$, $N=0$, gives, initially for $\Re s>1$ and then near $s=1$ after pole removal,

$$\zeta (s)=\frac{1}{s-1}+\frac{1}{2}-s{\int }_1^{\infty }{P}_1(t)t^{-s-1}dt.\text{\hspace{1em}(CS4)}$$

The integral and all fixed-order derivatives converge locally uniformly near one. Coefficient comparison with CS1 gives

$${\gamma }_k=\frac{1_{k=0}}{2}+{\int }_1^{\infty }\frac{P_1(t)}{t^2}[k(\log t{)}^{k-1}-(\log t{)}^k]dt,$$

where the first summand inside the brackets is zero for $k=0$. The finite binomial sum in CS1, the derivative identity DLMF 18.9.23, and the contiguous identity DLMF 18.9.13 therefore yield

$$S_{\gamma }(n)=-\frac{n}{2}+{\int }_0^{\infty }{P}_1(e^u)e^{-u}{L}_{n-1}^{(2)}(u)du.\text{\hspace{1em}(CS5)}$$

For $n=1$ this reads $S_{\gamma }(1)=-\gamma$, consistently with CS1. Put $p_n=L_{n-1}^{(2)}$. Telescoping the contiguous identity gives $p_n={\sum }_{k=0}^{n-1}(n-k)L_k^{(0)}$. Orthogonality against $L_0^{(0)}=1$ therefore gives ${\int }_0^{\infty }{e}^{-u}{p}_n(u)du=n$. Writing $F(u)=\{e^u\}=P_1(e^u)+1/2$ in CS5 yields the exact identity

$$U_n=n+S_{\gamma }(n)={\int }_0^{\infty }F(u)e^{-u}{p}_n(u)du.$$

Here is the standard coefficient budget used for this and the higher-power channel below. For measurable $g$ with

$$\mathcal{N}(g{)}^2:={\int }_0^{\infty }\frac{e^{-u}|g(u){|}^2}{u^2}du<\infty ,c_n(g)={\int }_0^{\infty }g(u)e^{-u}{p}_n(u)du,$$

work in $L^2((0,\infty ),u^2{e}^{-u}du)$ with $f=g/u^2$. The Laguerre row of DLMF Table 18.3.1 gives the orthonormal sequence $p_n/\sqrt{n(n+1)}$. Bessel’s inequality gives

$$\boxed{\sum _{n\ge 1}\frac{|c_n(g){|}^2}{n(n+1)}\le \mathcal{N}(g{)}^2.}\text{\hspace{1em}(CS6)}$$

For $F$, one has $F(u)=e^u-1=O(u)$ near zero and $0\le F(u)<1$ everywhere. Thus $\mathcal{N}(F)<\infty$, and CS6 proves

$$\boxed{\sum _{n\ge 1}\frac{|U_n{|}^2}{n(n+1)}\le \mathcal{N}(F{)}^2<\infty ,U_n=o(n),S_{\gamma }(n)=-n+o(n).}\text{\hspace{1em}(CS7)}$$

The little-$o$ conclusion follows because the summands tend to zero and $\sqrt{n(n+1)}/n\to 1$. No completeness or Parseval equality is needed. This supplies the reported $S_{\gamma }(n)=O(n)$ conclusion independently of the source’s displayed proof. It does not certify Corollary 16’s stronger $U_n=O(n^{1/4})$ rate or bound $P_n$.

CS4. Consumer: the original adjacent-Fibonacci joint matrix

Put $a=F_{j-1}$, $b=F_j$, $c=F_{j+1}=a+b$, for $j\ge 3$. Lagarias’s published Theorem 3.1, equations (3.3)–(3.4), supplies the actual Weil matrix on its Li test functions $G_b,G_c$:

$$K_j=\left(\begin{array}{cc}2{\lambda }_b& {\lambda }_b+{\lambda }_c-{\lambda }_a\\ {\lambda }_b+{\lambda }_c-{\lambda }_a& 2{\lambda }_c\end{array}\right).\text{\hspace{1em}(CS8)}$$

Its positivity is not supplied by the word “matrix” or “Weil”: without RH this Hermitian form is not known to be positive. Set

$$h_n=H_n+U_n,q_j=h_b+h_c-h_a,e_j=P_b+P_c-P_a,D_j=4h_b{h}_c-q_j^2.$$

Direct expansion of the same actual CS8, retaining every cross term, is the exact bookkeeping identity

$$\det K_j=D_j-E_j,E_j=4(h_c{P}_b+h_b{P}_c)-2q_j{e}_j+e_j^2-4P_b{P}_c,\text{\hspace{1em}(CS9)}$$

$$\mathrm{tr}{K}_j=2(h_b+h_c-P_b-P_c).\text{\hspace{1em}(CS10)}$$

Lagarias’s unconditional Theorem 5.1 supplies $H_n=\frac{1}{2}n\log n+\frac{1}{2}(\gamma -1-\log (2\pi ))n+O(1)$. Combining it with the sufficient CS7, and using that $a/b$ and $c/b$ stay bounded above and away from zero, gives

$$D_j=ab(\log b{)}^2+O(b^2\log b),2(h_b+h_c)=(b+c)\log b+O(b).\text{\hspace{1em}(CS11)}$$

Indeed $h_t=\frac{1}{2}t\log b+O(b)$ for $t\in \{a,b,c\}$; the leading terms in $4h_b{h}_c-q_j^2$ are $bc(\log b{)}^2-b^2(\log b{)}^2=ab(\log b{)}^2$. Both controlled baselines are eventually positive. The missing arithmetic estimates remain, jointly for every sufficiently large $j$,

$$\boxed{P_b+P_c\le h_b+h_c,E_j\le D_j.}\text{\hspace{1em}(CS12)}$$

Together they are equivalent to eventual positive semidefiniteness of the actual real symmetric CS8, by its trace and determinant. No bound for CS12 is proved here. Separate bounds must hold simultaneously for the same actual sequence and be quantitatively strong enough to imply both coupled inequalities. Separately attainable row configurations cannot be combined as one realization.

This controlled Stieltjes contribution is sufficient for CS11; it is not an upper bound on $P_n$, $E_j$, ${\lambda }_n$ or the Robin response. The Fibonacci absolute-growth bound, the original joint signed Robin estimate, the endpoint half-variance bound and RH remain unresolved.

CS5. The higher-prime-power channel has a finite Laguerre energy

The higher-prime-power estimates for compact autocorrelation tests in the weighted Mertens application retain their original test class and derivative norm. They are not substituted for a bound on the noncompact linear Laguerre row here. Instead, the standard Laguerre orthogonality already used in CS3 supplies a direct application to this actual row. No new prime-distribution theorem, priority claim, Lean certification or primary-prime estimate is asserted.

For $n\ge 1$, retain every proper prime power in

$$B_n=\sum _{p\in \mathbb{P}}\sum _{k\ge 2}\frac{\log p}{p^k}{L}_{n-1}^{(1)}(k\log p),A(u)=\psi (e^u)-\vartheta (e^u)=\sum _{\begin{array}{c}p\in \mathbb{P},k\ge 2\\ k\log p\le u\end{array}}\log p.\text{\hspace{1em}(CS13)}$$

Here $\mathbb{P}$ is the set of primes, $\psi (x)={\sum }_{m\le x}\Lambda (m)$ and $\vartheta (x)={\sum }_{p\le x}\log p$. Each proper prime power is counted once with its von Mangoldt weight. In particular $A(u)=0$ on $0\le u<\log 4$; the unit endpoint from CS2 is not inserted again.

The existing Rosser–Schoenfeld source, Theorem 13, equation (3.36), printed p.71, directly supplies

$$0\le \psi (x)-\vartheta (x)<\kappa \sqrt{x}(x>0),\kappa =\mathrm{1.42620.}$$

This bound includes the full proper-prime-power sum. It gives the independently defined finite budget

$$C_A^2:={\int }_0^{\infty }\frac{e^{-u}A(u{)}^2}{u^2}du\le \frac{{\kappa }^2}{\log 4}<\infty .\text{\hspace{1em}(CS14)}$$

There is no origin singularity because $A$ vanishes before $\log 4$; at infinity the integrand is bounded by ${\kappa }^2/u^2$. No estimate of the primary-prime discrepancy or assumption about zeta zeros is needed.

For fixed $n$, CS13 is absolutely convergent. Stieltjes integration by parts, with the boundary terms zero at both zero and infinity, uses

$$\frac{d}{du}\left(e^{-u}{L}_{n-1}^{(1)}(u)\right)=-e^{-u}{L}_{n-1}^{(2)}(u)$$

from the same DLMF derivative and contiguous identities used in CS3. Thus

$$B_n={\int }_0^{\infty }A(u)e^{-u}{L}_{n-1}^{(2)}(u)du.\text{\hspace{1em}(CS15)}$$

Since $C_A=\mathcal{N}(A)$, CS6 applies directly with $g=A$. Thus, for every finite partial sum and then its monotone limit,

$$\boxed{\sum _{n\ge 1}\frac{|B_n{|}^2}{n(n+1)}\le C_A^2<\infty .}\text{\hspace{1em}(CS16)}$$

In particular $|B_n|\le C_A\sqrt{n(n+1)}$ for every $n\ge 1$, and

$$\boxed{B_n=o(n).}\text{\hspace{1em}(CS17)}$$

No completeness assertion or Parseval equality is needed. This is a standard orthogonal coefficient estimate with its actual arithmetic input identified, not an estimate obtained by assigning the Robin slack to a norm.

CS6. The same joint matrix with a centered primary-prime row

Set $Q_n=P_n-B_n$ and $h_n^{\ast }=H_n+U_n-B_n$. Absolute convergence of CS13 permits the exact same-cutoff expression

$$Q_n=\underset{X\to \infty }{\lim }\left[\sum _{p\le X}\frac{\log p}{p}{L}_{n-1}^{(1)}(\log p)-\sum _{2\le m\le X}\frac{1}{m}{L}_{n-1}^{(1)}(\log m)\right].\text{\hspace{1em}(CS18)}$$

Both terms retain the common cutoff. Neither divergent summand is assigned a separate value. The actual Li identity is now ${\lambda }_n=h_n^{\ast }-Q_n$, with no prime power discarded.

The controlled combination also has a common-source budget, not just two separately chosen estimates:

$$\sum _{n\ge 1}\frac{|U_n-B_n{|}^2}{n(n+1)}\le \mathcal{N}(F-A{)}^2<\infty .$$

This is CS6 applied to the one actual function $F-A$; its norm is finite by the triangle inequality. Consequently $h_n^{\ast }=H_n+o(n)$.

For the same adjacent Fibonacci triple, the complete higher-power matrix contribution is

$${\mathcal{B}}_j=\left(\begin{array}{cc}2B_b& B_b+B_c-B_a\\ B_b+B_c-B_a& 2B_c\end{array}\right),\Vert {\mathcal{B}}_j{\Vert }_{\mathrm{op}}=o(b).\text{\hspace{1em}(CS19)}$$

Indeed all three indices are comparable and CS17 holds for every integer index. This is a bound on that component of the actual joint matrix, without a sign assertion about its entries. The baseline CS11 remains valid with $h$ replaced by $h^{\ast }$, since $B_t=o(t)$.

Equations CS9–CS10 apply exactly with $(h,P)$ replaced by $(h^{\ast },Q)$. Their coupled trace and determinant inequalities on this one actual centered prime sequence remain unproved. In particular every mixed $B$–$Q$ contribution is retained through $h^{\ast }$, rather than omitted from a determinant expansion. CS16–CS19 control the higher-power channel; they supply no growth bound for $Q_n$ or ${\lambda }_n$, no signed Robin estimate, and no proof of RH.

CS7. The critical primary-prime energy is an RH-level premise

The staircase of the actual same-cutoff row CS18 is

$$T(u)=\vartheta (e^u)-\lfloor e^u\rfloor +1,T(u)=0(0\le u<\log 2).$$

Its Stieltjes measure has weight $\log p$ at each primary prime and weight $-1$ at each integer at least two. Applying the derivative identity in CS5 to this single staircase gives, for each fixed $n$,

$$Q_n={\int }_0^{\infty }T(u)e^{-u}{p}_n(u)du=c_n(T).\text{\hspace{1em}(CS20)}$$

The lower boundary term is zero. The existing PNT supplier makes the upper boundary term tend to zero and the last integral absolutely convergent for each fixed polynomial $p_n$. This does not imply a uniform bound as $n$ grows. It preserves the common cutoff in CS18; neither divergent summand is separately assigned a value.

Define, for $\eta \in \{\vartheta ,\psi \}$,

$$J_{\eta }={\int }_2^{\infty }\frac{|\eta (x)-x{|}^2}{x^2(\log x{)}^2}dx.$$

The change of variable $x=e^u$ identifies $\mathcal{N}(T{)}^2$ with the same integral using $\vartheta (x)-\lfloor x\rfloor +1$ in the numerator. Its difference from $\vartheta (x)-x$ is $\{x\}+1$, a bounded function with finite weighted norm on $[2,\infty )$. The Rosser–Schoenfeld bound in CS5 also gives a finite weighted norm for $\psi -\vartheta$, since ${\int }_2^{\infty }dx/[x(\log x{)}^2]<\infty$. Thus

$$\mathcal{N}(T)<\infty ⟺J_{\vartheta }<\infty ⟺J_{\psi }<\infty .\text{\hspace{1em}(CS21)}$$

The classical Cramér square-mean estimate, printed p.147, supplies under RH $M(X):={\int }_2^X|\psi (t)-t{|}^2/tdt=O(X)$. For $k\ge 1$, its contribution to $J_{\psi }$ on $[2^k,2^{k+1}]$ is at most

$$\frac{M(2^{k+1})}{2^k(k\log 2{)}^2}=O(k^{-2}).$$

Summing proves $J_{\psi }<\infty$ under RH. The classical mean-square result is reused with its stated weight and hypothesis; the dyadic transfer is its parameter application to CS21.

Conversely, if $\mathcal{N}(T)<\infty$, the already established CS6 and CS20 give $Q_n=o(n)$. The exact CS3/CS18 split and CS7/CS17 then give ${\lambda }_n=H_n+o(n)=O(n\log n)$. This absolute growth implies RH by Lagarias’s published all-index criterion, Theorem 2.2(3), equation (2.22), already used in the Fibonacci growth application. No new completeness assertion, zero truncation or sign-sensitive Theorem 6.1 is needed. Consequently the exact strength of this proposed budget is

$$\boxed{\mathcal{N}(T)<\infty ⟺\mathrm{RH}.}\text{\hspace{1em}(CS22)}$$

This is a paper application of classical sources and the existing coefficient interface, without Lean certification or a priority claim. It classifies a possible sufficient route; no unconditional finite energy or new actual prime estimate has been acquired. Proving that energy finite remains a legitimate RH strategy, but assuming it does not supply an independently available reserve. The original signed trace/determinant conditions remain unpaid; a norm is not imposed as the necessary intermediate mechanism for any particular finite triple. The original Robin estimate and endpoint half-variance bound also remain unresolved.

CS8. A uniform arithmetic tail for growing Li tests

The following parameter application controls the omitted part of the actual ordered CS18 row simultaneously for growing degrees. It uses the existing PNT supplier and the classical finite Laguerre series; it is not a new prime-distribution estimate or a priority claim. The finite signed head, the original joint inequalities and RH remain unresolved. No numerical execution or practical cutoff claim is made.

For an integer $X\ge 3$, let $Q_n(X)$ be the bracket in CS18 before taking its limit, with the same cutoff in both sums, and put $U=\log X$. The exact finite-endpoint identity behind CS20 gives

$$Q_n-Q_n(X)={\int }_U^{\infty }T(u)e^{-u}{p}_n(u)du-e^{-U}{L}_{n-1}^{(1)}(U)T(U).\text{\hspace{1em}(CS23)}$$

In particular the upper boundary term is retained. The Johnston–Yang estimate, Theorem 1.1, and the Rosser–Schoenfeld complete higher-power bound already used in CS5 give, for $u\ge \log 2$,

$$\begin{array}{c}|T(u)|\le e^u\epsilon (u),\\ \epsilon (u)=9.39u^M{e}^{-d\sqrt{u}}+\kappa e^{-u/2}+2e^{-u},M=1.515,d=0.8274,\kappa =\mathrm{1.42620.}\end{array}$$

Indeed $T=(\psi (e^u)-e^u)-A(u)+\{e^u\}+1$. This uses the all-prime-power PNT statement and the full $A$, with no global RH premise or discarded prime powers.

Coffey’s equation (64), with $(\nu ,n)=(1,n)$ and $(2,n+1)$, is the classical terminating-series identity

$$L_{n-1}^{(\alpha )}(u)=\sum _{k=0}^{n-1}(-1{)}^k(\genfrac{}{}{0ex}{}{n-1+\alpha }{n-1-k})\frac{u^k}{k!},\alpha \in \{1,2\}.$$

For $u\ge 1$ and $1\le n\le N$, each $u^k/k!$ is at most $u^{N-1}$ and the binomial coefficient sum is at most $2^{N+1}$. Applying this bound to both terms of CS23 gives the explicit allowance

$$\boxed{\underset{1\le n\le N}{\sup }|Q_n-Q_n(X)|\le {\mathcal{E}}_N(U):=2^{N+1}\left[U^{N-1}\epsilon (U)+{\int }_U^{\infty }{u}^{N-1}\epsilon (u)du\right],U\ge 1.}\text{\hspace{1em}(CS24)}$$

All terms on the right refer to one common cutoff. This controls both signs of the omitted ordered tail, rather than assigning separate values to its divergent prime and integer sums.

One deterministic analytical cutoff is $X_N=\lceil \exp (N^4)\rceil$. There are constants $C_{\ast }>0$ and an integer $N_0$ such that, for every $N\ge N_0$ and every integer $X$ with $\log X\ge N^4$,

$${\mathcal{E}}_N(\log X)\le C_{\ast }{e}^{-\delta N^2},\delta =d/4>0.\text{\hspace{1em}(CS25)}$$

Here the constants are independent of $n,N,X$. To verify the uniform quantifier, completing squares gives, for $u\ge 1$,

$$\epsilon (u)\le C_{\epsilon }{u}^M{e}^{-d\sqrt{u}},C_{\epsilon }=9.39+\kappa e^{d^2/2}+2e^{d^2/4}.$$

Since $\log u/\sqrt{u}$ decreases on $[e^2,\infty )$, for all sufficiently large $N$ and every $u\ge N^4$,

$$\frac{(N+1)\log 2+(N-1+M)\log u}{\sqrt{u}}\le \frac{(N+1)\log 2+4(N-1+M)\log N}{N^2}\le d/2.$$

Thus the integrand and endpoint in CS24 are bounded by $C_{\epsilon }{e}^{-d\sqrt{u}/2}$, giving

$${\mathcal{E}}_N(U)\le C_{\epsilon }\left(1+\frac{4\sqrt{U}}{d}+\frac{8}{d^2}\right)e^{-d\sqrt{U}/2}.$$

Absorbing this polynomial factor into $e^{d\sqrt{U}/4}$ proves CS25. The threshold is an analytical existence statement; no scan, numerical optimization or efficient computation is asserted.

For the actual adjacent Fibonacci triple $a,b,c$, set $N=c$ and use the common integer $X_c$. Define ${\lambda }_n^{(X)}=h_n^{\ast }-Q_n(X)$ for $n\in \{a,b,c\}$, retaining the actual full $h^{\ast }=H+U-B$. Let $K_j^{(X)}$ be CS8 with these three values substituted. With ${ϵ}_c=C_{\ast }{e}^{-\delta c^2}$, the diagonal errors are at most $2{ϵ}_c$ and the off-diagonal error at most $3{ϵ}_c$. The maximum absolute row sum therefore gives the whole real symmetric matrix allowance

$$\boxed{\Vert K_j-K_j^{(X_c)}{\Vert }_{\mathrm{op}}\le 5{ϵ}_c}\text{for every sufficiently large }j.\text{\hspace{1em}(CS26)}$$

Every mixed contribution and the unit endpoint remain in the original exact split. This upgrades fixed-degree convergence to a simultaneous bound for this growing family. It supplies no eventual sign for the finite-head matrices $K_j^{(X_c)}$; that common-source signed comparison is still required. Finite truncation with a small omitted-tail allowance is not a verification over all Fibonacci indices, a Robin estimate, the endpoint half-variance bound, or a proof of RH. This application has no new Lean certification.

CS9. A classical bound pays the growing low-prime prefix

The classical uniform Laguerre bound as reproduced in Plewa, Section 4, equation (19), applies to the actual finite CS18 row with the same inclusive integer cutoff. This is a parameter application of that existing estimate, not a new polynomial or prime-distribution theorem, numerical validation or priority claim.

There are constants $C_0,N_0$ such that for every integer $N\ge N_0$ and every integer $N/3\le n\le N$,

$$\boxed{|Q_n(N)|\le {\eta }_N:=C_0{N}^{3/4}(\log N{)}^{1/4}=o(N).}\text{\hspace{1em}(CS27)}$$

To apply the source, its standard function at degree $n-1$ and parameter $1$ is exactly ${\mathcal{L}}_{n-1}^1(u)=\sqrt{u/n}{e}^{-u/2}{L}_{n-1}^{(1)}(u)$. The cited estimate, with $\nu =4n$, therefore gives

$$|L_{n-1}^{(1)}(u)|\le C{n}^{1/4}{e}^{u/2}{u}^{-3/4},(4n{)}^{-1}<u\le 2n.$$

Its constant is independent of $n,u$. For all sufficiently large $N$, $(4n{)}^{-1}<\log 2$ and $\log N\le 2n$ hold throughout $N/3\le n\le N$. Consequently the bound holds on the entire growing interval $\log 2\le u\le \log N$ needed by both finite sums. It is not an application of a fixed-$u$ asymptotic with uncontrolled constants. Using only that primes form a subset of the integers gives

$$\begin{array}{rl}\left|\sum _{p\le N}\frac{\log p}{p}{L}_{n-1}^{(1)}(\log p)\right|& \le C{n}^{1/4}(\log N{)}^{1/4}\sum _{2\le m\le N}{m}^{-1/2}\le 2C{N}^{3/4}(\log N{)}^{1/4},\\ \left|\sum _{2\le m\le N}\frac{1}{m}{L}_{n-1}^{(1)}(\log m)\right|& \le C{n}^{1/4}(\log 2{)}^{-3/4}\sum _{2\le m\le N}{m}^{-1/2}\le 2C(\log 2{)}^{-3/4}{N}^{3/4}.\end{array}$$

The triangle inequality proves CS27 after absorbing the fixed second constant. Indeed ${\eta }_N/N=C_0(\log N/N{)}^{1/4}\to 0$. No cancellation or prime-counting theorem is used. Both sums keep $p\le N$ and $2\le m\le N$ exactly, and the unit endpoint is not moved.

For the original adjacent Fibonacci triple $a,b,c$, all three indices lie in $[c/3,c]$. Keep CS25’s single integer $X_c=\lceil \exp (c^4)\rceil$ and define the actual remaining middle by

$$Q_n^{\mathrm{mid}}=Q_n(X_c)-Q_n(c),{\lambda }_n^{\mathrm{mid}}=h_n^{\ast }-Q_n^{\mathrm{mid}},n\in \{a,b,c\}.$$

Thus the middle sums have $c<p\le X_c$ and $c<m\le X_c$; $h^{\ast }=H+U-B$ remains whole. In particular all proper prime powers and the unit endpoint remain in their existing channels. With $q_n=Q_n(c)$ and $r_n=Q_n-Q_n(X_c)$, the exact relation is ${\lambda }_n^{\mathrm{mid}}-{\lambda }_n=q_n+r_n$. Let $\mathcal{M}(z_a,z_b,z_c)$ denote CS8’s linear matrix map. Then

$$K_j^{\mathrm{mid}}-K_j=\mathcal{M}(q_a+r_a,q_b+r_b,q_c+r_c),$$

with the original off-diagonal combination retained. Applying the same row-sum allowance as CS26 gives, for every sufficiently large Fibonacci index,

$$\boxed{\Vert K_j^{\mathrm{mid}}-K_j{\Vert }_{\mathrm{op}}\le 5\left({\eta }_c+C_{\ast }{e}^{-\delta c^2}\right)=o(c).}\text{\hspace{1em}(CS28)}$$

This pays the actual growing low-prime prefix and carries the already paid remote tail, leaving the actual middle interval in the original signed matrix. It does not bound the sign of that middle or establish its eventual positivity. The original complete joint estimate, Robin, the endpoint half-variance bound and RH remain unresolved. This application has no new Lean certification.

CS10. Two weaker signed targets on the actual middle

Keep the actual middle and the full $h^{\ast }$ from CS9. Instead of requiring positivity of the whole $K_j^{\mathrm{mid}}$, consider the following pair of unproved estimates: there is one $C>0$ such that for every sufficiently large adjacent Fibonacci triple $a,b,c$,

$$\boxed{{\lambda }_c^{\mathrm{mid}}\ge -C{c}^2,2{\lambda }_a^{\mathrm{mid}}+2{\lambda }_b^{\mathrm{mid}}-{\lambda }_c^{\mathrm{mid}}\ge -C{c}^2.}\text{\hspace{1em}(CS29)}$$

The first scalar is half the second diagonal entry of $K_j^{\mathrm{mid}}$; the second is $(2,-1)K_j^{\mathrm{mid}}(2,-1{)}^{\mathsf{T}}/2$. Thus positive semidefiniteness would imply both. Their exact arithmetic requirements, keeping every term of $h^{\ast }$, are

$$\begin{array}{rl}{Q}_c^{\mathrm{mid}}& \le h_c^{\ast }+C{c}^2,\\ 2Q_a^{\mathrm{mid}}+2Q_b^{\mathrm{mid}}-Q_c^{\mathrm{mid}}& \le 2h_a^{\ast }+2h_b^{\ast }-h_c^{\ast }+C{c}^2.\end{array}\text{\hspace{1em}(CS30)}$$

CS28 transfers each scalar to the actual Li coefficients with $o(c)$ error. Increasing a constant would therefore give ${\lambda }_c\ge -C^{\prime }{c}^2$ and ${\lambda }_c\le 2{\lambda }_a+2{\lambda }_b+C^{\prime }{c}^2$, whence

$$|{\lambda }_c|\le 2|{\lambda }_a|+2|{\lambda }_b|+C^{\prime }{c}^2.$$

Since $a/c\in [1/3,1/2]$, $2a^3+2b^3=2c^3[1-3(a/c)+3(a/c{)}^2]\le 2c^3/3$. Choose $D\ge 3C^{\prime }$ large enough to cover the two Fibonacci seeds before the eventual recurrence begins. Induction would then give $|{\lambda }_{F_j}|\le D{F}_j^3$: the recurrence is at most $2D{c}^3/3+C^{\prime }{c}^2\le D{c}^3$. The existing Fibonacci absolute-growth criterion would imply RH.

This is a sufficient unpaid target, not an estimate of the actual middle or a new unconditional RH result. Deriving another sufficient condition alone does not supply arithmetic cancellation. The argument is a paper application of the existing matrix transport and growth criterion, without Lean certification or a priority claim.

CS11. The prescribed absolute PNT allowance cannot pay that target

Keep CS8’s $\epsilon$, $M=1.515$ and $d=0.8274$, and CS7’s $p_n(u)=L_{n-1}^{(2)}(u)$. For integers $3\le L<X$, the same finite Stieltjes integration by parts underlying CS23 gives, with $k_n(u)=e^{-u}{L}_{n-1}^{(1)}(u)$,

$$\begin{array}{rl}{Q}_n(X)-Q_n(L)=& {\int }_{\log L}^{\log X}T(u)e^{-u}{p}_n(u)du\\ & +k_n(\log X)T(\log X)-k_n(\log L)T(\log L).\end{array}$$

Both endpoints remain. For the second CS30 combination define

$$J_c(u)=2p_a(u)+2p_b(u)-p_c(u),{\mathcal{A}}_j={\int }_{\log c}^{\log X_c}\epsilon (u)|J_c(u)|du.\text{\hspace{1em}(CS31)}$$

This is an absolute comparison allowance, obtained by replacing $e^{-u}|T(u)|$ with its CS8 upper bound. It is not the actual signed integral or its absolute value. Adding the absolute values of the two joint endpoint terms only increases this allowance. Even this integral part satisfies

$$\boxed{{\mathcal{A}}_j^{1/c}⟶\infty .}\text{\hspace{1em}(CS32)}$$

To apply the classical terminating Laguerre series already cited in CS8, take the single analytical interval $I_c=[4c^2,4c^2+1]$. It is contained in $[\log c,\log X_c]$ for all sufficiently large $c$. For $1\le r\le n-1$, the absolute ratio of the $r$-th subleading term of $p_n$ to $u^{n-1}/(n-1)!$ is

$$\left(\genfrac{}{}{0ex}{}{n+1}{r}\right)\frac{(n-1)!}{(n-1-r)!}{u}^{-r}\le \frac{1}{r!}{\left(\frac{n^2-1}{u}\right)}^r\le 4^{-r},n\le c,u\ge 4c^2.$$

Their sum is at most ${\sum }_{r\ge 1}{4}^{-r}=1/3$. Thus

$$\frac{2}{3}\frac{u^{n-1}}{(n-1)!}\le |p_n(u)|\le \frac{4}{3}\frac{u^{n-1}}{(n-1)!},1\le n\le c,u\ge 4c^2.\text{\hspace{1em}(CS33)}$$

For $n=1$ the subleading sum is empty. Taking ratios gives $|p_b|/|p_c|\le 2(c/u{)}^{c-b}\le 2(4c{)}^{-a}$ and $|p_a|/|p_c|\le 2(4c{)}^{-b}$ on $I_c$. As $a,b\to \infty$, the triangle inequality therefore gives $|J_c|\ge |p_c|/2\ge u^{c-1}/[3(c-1)!]$ there for every sufficiently large Fibonacci index. The two lower degrees cannot cancel the highest degree on this range. Using just the first positive summand of $\epsilon$ and $\sqrt{4c^2+1}\le 2c+1$ yields

$${\mathcal{A}}_j\ge \frac{9.39}{3}\frac{(4c^2{)}^{c-1+M}}{(c-1)!}{e}^{-d(2c+1)}.\text{\hspace{1em}(CS34)}$$

Replacing $(c-1)!$ by $c^{c-1}$ in the denominator of CS34 gives a coarser lower bound. The $c$-th root of that coarser bound divided by $c$ has limit $4e^{-2d}>0$, proving CS32. This is a finite-series parameter application on one fixed analytical interval, with no numerical execution, cutoff optimization, new Laguerre theorem or literature-priority claim.

Consequently this particular absolute PNT comparison allowance cannot have a polynomial or subexponential upper bound in $c$, even when the Fibonacci combination is formed before taking its absolute value. This does not lower-bound actual $T$, $Q_n^{\mathrm{mid}}$, a Li coefficient or the signed joint form. It does not rule out further signed arithmetic information, exact endpoint cancellation or a different representation. What remains unpaid is an estimate of the actual same-source middle with both endpoints, sufficient for CS29 or another valid RH route. Robin, the complete joint estimate, endpoint half-variance and RH remain unresolved; no new Lean certification is asserted.

CS12. A positive-data representation with a signed inversion

The generalized Li interface supplies another exact representation of the same actual critical coefficients. Its common finite transfer TL3 uses the actual nonnegative $\tau =2$ coefficients ${\Lambda }_m(2)$, with one prescribed regularization and integer cutoff for every $1\le n\le c$. The uniform approximation in TL4 gives the same original Fibonacci matrix a whole-matrix error $5C{e}^{-c^2/2}$ for all sufficiently large indices. This is an approximation-error estimate, not a sign or growth bound on the critical Li coefficients or the actual arithmetic middle.

At $\tau =2$ a classical Poisson representation gives an unconditional positive semidefinite comparison matrix. Returning to $\tau =1$ uses signed binomial weights: the coefficient matrix already at index $m=a<b<c$ is indefinite. Thus no termwise positivity transport is supplied. The same-source weighted sum retains the original unresolved sign problem; positivity of the whole transferred matrix is not asserted. The CS29 targets, Robin, endpoint half-variance and RH remain unproved. The source identities and estimates are directly reused; the growing-parameter transfer is a paper application without Lean certification, numerical execution or a literature-priority claim.

CS13. An explicit whole-matrix transfer and its finite signed consumer

This application makes the original CS28 approximation effective, with explicit constants and a specified starting scale. The same comparison also supplies a finite positive bound for the original arithmetic middle. It reuses the classical Laguerre equation and energy comparison, global prime count, PNT tail allowance, reflected zero matrices and published finite-height verification. These are paper applications without new Lean certification or a literature-priority claim.

Exact source and target

Keep CS18, CS23–CS24, CS28’s exact arithmetic definition and original matrix map. For adjacent positive Fibonacci $a\le b$, $c=a+b$, use $X_c=\lceil \exp (c^4)\rceil$ and $Q_n^{mid}=Q_n(X_c)-Q_n(c)$, ${\lambda }_n^{mid}=h_n^{\ast }-Q_n^{mid}$, $n\in \{a,b,c\}$, with the complete $h^{\ast }=H+U-B$, every proper prime power and both original inclusive cutoffs. First make the original approximation effective at every such triple with $c\ge 10^{24}$, then consume it in a finite all-vector bound:

$$K_j^{mid}⪰c{I}_2/50000(10^{24}\le c\le 5\cdot 10^{25}).$$

The matrix is the ORIGINAL actual arithmetic middle, not a new surrogate or a scalar Li-positivity inference. The full RH/strictRobin target remains unchanged and unproved.

An explicit finite comparison from the classical equation

The standard uniform Laguerre estimate already supplies the shape of the prefix bound, and is reused as background rather than presented as new. For a finite constant use the classical differential equation in DLMF Table 18.8.1, its Laguerre row, with degree $n-1$ and parameter $1$. Together with the usual Liouville change of variable and energy/Gronwall estimate it permits the following explicit parameter comparison, without any unknown asymptotic start.

Write $L=\log c$, $x_0=1/n$ and

$$Y(x)=x{e}^{-x/2}{L}_{n-1}^{(1)}(x),p(x)=n/x-1/4.$$

The cited equation gives $Y^{\prime \prime }+pY=0$. For every $c\ge 10^{24}$, $(\log c)/c$ decreases and $\log (10^{24})<60$, so $L<c/3\le n$. Throughout $x_0\le x\le L$, therefore $x\le n$ and $p\ge 3n/(4x)>0$. The terminating series already used in CS8 has decreasing alternating terms at $x_0$. Its first two terms and the derivative identity $(L_{n-1}^{(1)}{)}^{\prime }=-L_{n-2}^{(2)}$ give

$$n/2\le L_{n-1}^{(1)}(x_0)\le n,-n(n-1)/2\le (L_{n-1}^{(1)}{)}^{\prime }(x_0)\le 0.$$

Thus $0<Y(x_0)\le 1$ and $0\le Y^{\prime }(x_0)\le n$; for the lower derivative bound the bracket is at least $(1-1/(2n))n/2-(n-1)/2=1/4$.

Use the classical change $dt/dx=\sqrt{p}$, $V(t)=p(x{)}^{1/4}Y(x)$. Its exact transformed equation is

$$V_{tt}+(1+r)V=0,r=\frac{5p^{\prime 2}-4p{p}^{\prime \prime }}{16p^3}=\frac{n(2x-3n)}{16x(n-x/4{)}^3}.$$

Since $x\le n$, $|r|\le 4/(9nx)$, and hence

$${\int }_{x_0}^L|r|\sqrt{p}dx\le \frac{4}{9\sqrt{n}}{\int }_{1/n}^L{x}^{-3/2}dx<8/9.$$

At $x_0$, $3n^2/4\le p(x_0)<n^2$ and $|p^{\prime }(x_0)|=n^3$. The preceding initial bounds give

$$|V_0|\le \sqrt{n},|(V_t{)}_0|\le \left((4/3{)}^{1/4}+\frac{1}{4}(4/3{)}^{5/4}\right)\sqrt{n}<\frac{16}{9}\sqrt{n}.$$

The ordinary energy $E=V^2+V_t^2$ satisfies $|dE/dt|\le |r|E$. Integrating this inequality gives

$$|V|\le \sqrt{E}\le \frac{25}{9}\sqrt{n}\exp (4/9)<\frac{50}{9}\sqrt{n}.$$

Here ${\sqrt{E}}_0\le |V_0|+|(V_t{)}_0|$ and $\exp (4/9)<2$. Returning to the original polynomial and using $p^{-1/4}\le (4/3{)}^{1/4}(x/n{)}^{1/4}<(4/3)(x/n{)}^{1/4}$ yields

$$|L_{n-1}^{(1)}(u)|\le \frac{200}{27}{n}^{1/4}{e}^{u/2}{u}^{-3/4}<10n^{1/4}{e}^{u/2}{u}^{-3/4},\log 2\le u\le L,c/3\le n\le c.\text{\hspace{1em}(EB1)}$$

This comparison holds for every $c\ge 10^{24}$. The integration begins at $1/n\le \log 2$, so the actual finite prefix is contained in the proven range. These are explicit parameter calculations from a classical equation and energy comparison for the required consumer, with no claim of a new general Laguerre inequality.

Finite low prefix using the existing global prime count

For every $c\ge 10^{24}$, put $L=\log c>48$. Let $w(t)=t^{-1/2}(\log t{)}^{-3/4}$ and $g(t)=\sqrt{t}(\log t{)}^{-3/4}$. For $t\ge e^3$, $g^{\prime }(t)=(1/2-3/(4\log t))w(t)\ge w(t)/4$. Also $w(2)<2$. The function $g$ increases on $[e^3,\infty )$, so $g(c)\ge g(10^{24})>10^{12}/60>100$. Thus

$${\int }_2^{c}w(t)dt<54+4g(c)<5g(c).\text{\hspace{1em}(EB2)}$$

The lower interval is bounded by $e^{3}w(2)<27\cdot 2=54$. Reuse Rosser–Schoenfeld Corollary1 eq(3.6), already owned in the existing Rosser–Schoenfeld entry: $\pi (x)<1.25506x/\log x$ for $x>1$. It directly gives $\vartheta (x)\le \pi (x)\log x<2x$. Stieltjes partial summation, with $\vartheta (2^{-})=0$, gives

$$\sum _{p\le c}\log pw(p)\le 2cw(c)+2{\int }_2^{c}t(-w^{\prime }(t))dt=4w(2)+2{\int }_2^{c}w(t)dt<11g(c).$$

Monotonicity similarly gives ${\sum }_{2\le m\le c}w(m)\le w(2)+{\int }_2^{c}w(t)dt<6g(c)$. These are upper bounds on positive envelopes of the two ACTUAL finite sums; no primary-prime cancellation is assumed. EB1 then gives for all three actual indices

$$|q_n|=|Q_n(c)|\le 200c^{3/4}{L}^{-3/4}.\text{\hspace{1em}(EB3)}$$

The constant 200 rounds $10(11+6)=170$ outward. This replaces the prime-subset-of-integers log loss by the already available global count; no new prime-count theorem is derived.

The unchanged remote cutoff has an explicit finite allowance

Use the complete CS24 formula with $N=c$, $U=\log X_c\ge c^4$. Its existing constants satisfy $C_{\epsilon }=9.39+\kappa e^{d^2/2}+2e^{d^2/4}<22$, $M=1.515<2$, $4/5<d=0.8274<1$ and $\kappa =1.42620<3/2$. Together with $e<3$ these give $\epsilon (u)\le 22u^M{e}^{-d\sqrt{u}}$. For every $u\ge c^4$, monotonicity of $(\log u)/\sqrt{u}$ on $[e^2,\infty )$ gives

$$\frac{(c+1)\log 2+(c-1+M)\log u}{\sqrt{u}}\le \frac{2(1+4\log c)}{c}<\frac{482}{10^{24}}<\frac{2}{5}.$$

The last comparison uses that $(1+4\log c)/c$ decreases for $c\ge 10^{24}$ and $\log (10^{24})<60$. Consequently the endpoint and integrand in the owned CS24 are at most $22e^{-2\sqrt{u}/5}$, and direct substitution into its displayed integral bound gives

$$|r_n|=|Q_n-Q_n(X_c)|\le 22(14+5\sqrt{U})e^{-2\sqrt{U}/5}<1.\text{\hspace{1em}(EB4)}$$

For the last outward comparison put $z=\sqrt{U}\ge c^2\ge 10^{48}$. Since $14+5z\le 6z$ and $\exp (2z/5)\ge (2z/5{)}^3/6$, the bound is at most $12375/z^2<1$. No unknown asymptotic start, prime enumeration, exponential evaluation or numerical experiment is substituted for CS24. Every actual endpoint term remains.

The exact original CS28 relation and row sum now give

$$\Vert K_j^{mid}-K_{a,b}{\Vert }_{op}\le 1000c^{3/4}(\log c{)}^{-3/4}+5<c/16000+5,c\ge 10^{24}.\text{\hspace{1em}(EB5)}$$

Here $c^{1/4}\ge 10^6$ and $L^{3/4}>16$ because $48^3>16^4$.

Pay the finite comparison with the same complete zero matrix

Now restrict $10^{24}\le c\le 5\cdot 10^{25}$.

Use the exact Gram determinant/trace and reflected error in PTL3–PTL4 directly. For $\gamma >c$, $\theta \ge \gamma /({\gamma }^2+1)\ge 5/(6\gamma )$, and $\sin x\ge 5x/6$ for $0\le x\le 1/2$. Substitution into that exact formula and $a\ge c/3$, $b\ge c/2$, $b^2+c^2\le 2c^2$ gives

$${\lambda }_{min}(G_{\rho })\ge (5/6{)}^{10}\frac{c^4}{144{\gamma }^4}>\frac{c^4}{1152{\gamma }^4}>\frac{c^4}{1536{\gamma }^4}.$$

Indeed $(5/6{)}^5=3125/7776>2/5$, so its tenth power exceeds $4/25>1/8$. The existing radial row allowance is $2c^2/{\gamma }^4$, which is at most $c^4/(6144{\gamma }^4)$ for $c^2\ge 12288$. Hence the SAME complete tail obeys $K^{tail}(c)⪰c^4{S}_4^{+}(c)I/2048$. Only this application’s outward constants change; no standalone constant-improvement result or repeated generic proof is retained.