bibkey: erdosnicolas1975repartition authors: Paul Erdős; Jean-Louis Nicolas year: 1975 title: Répartition des nombres superabondants doi: 10.24033/bsmf.1793 url: https://www.numdam.org/item/BSMF_1975__103__65_0/ claim: The classical benefit is the nonnegative loss relative to a colossally abundant optimizer; the FIB price-loss expression is exactly this existing quantity. strata_touched: [] license: citation-only triage: anchor

Répartition des nombres superabondants

Primary source: Bulletin de la Société Mathématique de France 103 (1975), 65–90, Numdam article record and original scan. The relevant locators are Proposition 4(c)–(d) and its proof on printed pp.70–71, and §3, Proposition 5 and its proof on pp.73–74. This note records the definitions and their scope; it is not a verification of every proof in the paper or a Lean result.

Let $Z(n)=\sigma (n)/n$, and let $N$ maximize $Z(n)n^{-ϵ}$ for a fixed $ϵ>0$. On printed p.74, the authors define, for an arbitrary positive integer $m$,

$${\mathrm{ben}}_{N,ϵ}(m)=ϵ\log (m/N)-\log \frac{Z(m)}{Z(N)}\ge 0.$$

The nonnegativity follows from the defining optimality of the same $N$ and $ϵ$. If $L=\log m$, $F=\log Z(m)$, $L_0=\log N$, $F_0=\log Z(N)$, then

$$F_0-ϵL_0-F+ϵL={\mathrm{ben}}_{N,ϵ}(m).$$

Thus a “price loss” with the same reference optimizer and price is exactly the classical benefit, not a new FIB invariant. In particular, a separately justified choice $N=5040$, $ϵ=1/25$ has this meaning; this card does not independently establish that optimizer choice.

Proposition 5 is a near-extremal structural application. Its additional hypotheses are that $n$ is superabundant and it lies between $N$ and $NP$, where $P$ is the prime after the largest prime factor of $N$. Its proof uses a small benefit to control displacement of prime-exponent thresholds. The arbitrary-integer definition of benefit must not be confused with those stronger hypotheses on the integer to which Proposition 5 applies. An arbitrary Robin violation in a prescribed residue class is not automatically superabundant.

The proof points back to Proposition 4, p.120, of Nicolas’s Répartition des nombres hautement composés de Ramanujan, Canadian Journal of Mathematics 23 (1971), 116–130. Its original benefit decomposition and quadratic threshold-cost calculations on pp.117–120 have been inspected. That source uses the divisor-count objective and assumes a highly composite target for Proposition 4; the 1975 transfer uses the divisor-sum ratio and a superabundant target. Neither structural conclusion applies to an arbitrary host merely because its benefit is defined.

For the FIB Robin analysis, the valuation-increment source

$$w_s(d)=\frac{b_s(d)}{d},b_s(p^j)=Z(p^j{)}^s-Z(p^{j-1}{)}^s,J_s(d)=\log \frac{{\max }_e{w}_s(e)}{w_s(d)}$$

has a different objective. The classical support-loss method motivates its decomposition, but Proposition 5 supplies neither its uniform editing bound nor its power-sum or complementary-moment estimates without an additional argument. Conversely, the unweighted assertion that a growing reduced residue class has at most one sufficiently large Robin violation follows by a short classical support-loss argument applied to $n/\phi (n)$, combined with prime-number and Mertens estimates. That synthesis requires no Fibonacci encoding and must not be advertised as a FIB-specific method or historical novelty. The project’s weighted residual statement remains a separate assertion.

A finite resource certificate for the same arbitrary host

This application uses the classical benefit with its arbitrary-integer domain, finite layer-cake summation, and the quantitative prime estimates already cited in Dusart 2010. It does not apply Proposition 5’s superabundant conclusion to the host. The following is a paper derivation of a sufficient certificate, not a claim of a new benefit method or a Lean-verified estimate.

Fix $s>0$ and choose the maximal CA optimizer $C=C_s$ of $Z(m)m^{-1/s}$: include every zero-cost threshold layer. For the same actual integer $N\ge 1$, put $a_p=v_p(C)$, $e_p=v_p(N)$ and

$${\ell }_{p,j}=\log \frac{Z(p^j)}{Z(p^{j-1})},c_{p,j}=s{\ell }_{p,j}-\log p,D_s(N)=s{\mathrm{Ben}}_{C,1/s}(N).$$

The accepted layers are exactly $j\le a_p$, with $c_{p,j}\ge 0$; all later layers have $c_{p,j}<0$. Telescoping the existing benefit over the actually crossed layers gives

$$\begin{array}{rl}{D}_s(N)& =\sum _{p:e_p>a_p}\sum _{j=a_p+1}^{e_p}|c_{p,j}|+\sum _{p:e_p<a_p}\sum _{j=e_p+1}^{a_p}|c_{p,j}|,\\ E_{+}& =\sum _p(e_p-a_p{)}_{+}\log p,E_{-}=\sum _p(a_p-e_p{)}_{+}\log p,\\ E_{+}-E_{-}& =\log (N/C).\end{array}$$

Thus $E_{+}=\log u$ and $E_{-}=\log v$ for the reduced whole-host ratio $N/C=u/v$. In particular, with $g=\gcd (N,C_s)$,

$$E_{+}=\log (N/g),E_{-}=\log (C_s/g).$$

Once the same-price reference and the actual host have been obtained, these two resource inputs need only their gcd; factoring the host’s remaining cofactor is unnecessary for this lower certificate. This reference is $C_s$, not the increment reference $n_y$ in FIB equation (230.4); a reconstructed ratio to $n_y$ cannot silently replace it. No exponent rearrangement or replacement of $N$ is allowed.

For $0\le h<1$, define the available weak-cost resources

$$\begin{array}{rl}{A}_{+}(h)& =\sum _p\sum _{\begin{array}{c}j>a_p\\ |c_{p,j}|\le h\log p\end{array}}\log p,\\ A_{-}(h)& =\sum _p\sum _{\begin{array}{c}1\le j\le a_p\\ |c_{p,j}|\le h\log p\end{array}}\log p.\end{array}$$

Both are finite. For each side, the actual crossed resource costing more than $h$ per unit is at least $(E_{\pm }-A_{\pm }(h){)}_{+}$. The finite layer-cake formula therefore supplies, for $0<h_0<1$,

$$D_s(N)\ge {\int }_0^{h_0}(E_{+}-A_{+}(h){)}_{+}dh+{\int }_0^{h_0}(E_{-}-A_{-}(h){)}_{+}dh.$$

The two terms refer to the same exponent vector; they are not independently chosen optimal hosts.

Prime thresholds, higher layers, and endpoint costs

For $t>0$, let $x(t)>1$ be the unique root

$$t\log (1+1/x(t))=\log x(t).$$

Write $x_s=x(s)$, $x_{-}=x(s/(1+h))$ and $x_{+}=x(s/(1-h))$. The first-layer part of $A_{-}$ is exactly the logarithmic prime mass in the closed interval $[x_{-},x_s]$; the first-layer part of $A_{+}$ is that in $(x_s,x_{+}]$. In particular, deleting a tied first layer has zero cost and belongs to $A_{-}(0)$. The left endpoint must not be subtracted with the convention $\vartheta (x)={\sum }_{p\le x}\log p$ when it is prime.

For higher layers use the existing geometric marginal bound ${\ell }_{p,j}\le p^{-j}$; its repository statement is golden_layer_marginal_le_inv_pow in D5/S3/Arith/GoldenLayerMarginalDecay.lean, after multiplication by $\log p$. There is also a direct separation check: with $T_j=p+\cdots +p^j$,

$${\ell }_{p,j}=\log (1+1/T_j)>\frac{1}{T_j+1},p{\ell }_{p,j+1}<\frac{p}{T_{j+1}}=\frac{1}{T_j+1}.$$

Hence ${\ell }_{p,j}/{\ell }_{p,j+1}>p\ge 2$. For $h\le 1/3$, at most one layer per prime can have $s{\ell }_{p,j}/\log p\in [1-h,1+h]$. Every weak layer $j\ge 2$ also satisfies

$$p^2\log p\le \frac{s}{1-h}\le \frac{3s}{2}.$$

Let $P_2>1$ solve $P_2^2\log P_2=3s/2$. The entire higher-layer resource on either side is consequently at most $\vartheta (P_2)$, counting possible ties as well as nonzero costs.

Take $y\ge 6$, $s=y\log y$ and $0<h_0\le 1/3$. Set

$${\eta }_y=\underset{y/2\le u\le 3y/2}{\sup }|\vartheta (u)-u|,B_y=\vartheta (P_2)+2{\eta }_y+\log (3y/2).$$

The roots used for $0\le h\le h_0$ lie in $[y/2,3y/2]$. Indeed, $x_s<y$, and the function $T(x)=\log x/\log (1+1/x)$ obeys $T^{\prime }(x)>T(x)/x$ for $x>1$. This follows from

$$\frac{x{T}^{\prime }(x)}{T(x)}=\frac{1}{\log x}+\frac{1}{(x+1)\log (1+1/x)}\ge \frac{1}{\log x}+\frac{x}{x+1}>1.$$

It gives $x_{+}\le x_s/(1-h)$ and $x_{-}\ge x_s/(1+h)$. The remaining lower endpoint follows from $T(y/2)\le (y/2+1)\log (y/2)<3s/4$. Combining the interval masses, both prime errors and the higher layers gives the simultaneous finite bounds

$$\boxed{A_{+}(h)\le \frac{yh}{1-h}+B_y,A_{-}(h)\le \frac{yh}{1+h}+B_y(0\le h\le h_0).}$$

The $\log (3y/2)$ term pays the possible closed left endpoint. The cited prime estimates give $B_y=O(y/{\log }^{2}y+\sqrt{y})$, since $P_2<2\sqrt{y}$. These are bounds for every host at the same price; they do not assert a favorable sign for the prime error.

For a fixed $h>0$, the first-layer prime masses are asymptotic to $yh/(1-h)$ and $yh/(1+h)$ respectively. Thus a bound $A_{+}(h)\le (1+o(1))yh+o(y)$ cannot hold uniformly on a fixed nonzero $h$ interval. A common linear bound can use $yh/(1-h_0)+B_y$; its coefficient approaches one only when $h_0\to 0$.

The source-preserving Robin consumer

Put $z_{\pm }=(E_{\pm }-B_y{)}_{+}$ and define

$$\begin{array}{rl}{t}_{+}(z)& =\min \{h_0,z/(y+z)\},\\ t_{-}(z)& =\{\begin{array}{ll}\min \{h_0,z/(y-z)\},& z<y,\\ h_0,& z\ge y,\end{array}\\ G_{+}(z)& =(z+y)t_{+}(z)+y\log (1-t_{+}(z)),\\ G_{-}(z)& =(z-y)t_{-}(z)+y\log (1+t_{-}(z)).\end{array}$$

Integration of the two resource bounds gives $D_s(N)\ge G_{+}(z_{+})+G_{-}(z_{-})$. When $z_{+}\le y{h}_0/(1-h_0)$ and $z_{-}\le y{h}_0/(1+h_0)$, these simplify to

$$G_{+}(z_{+})=z_{+}-y\log (1+z_{+}/y),G_{-}(z_{-})=-z_{-}-y\log (1-z_{-}/y).$$

In the unsaturated ranges displayed above, for $z_{\pm }=o(y)$ each has leading term $z_{\pm }^2/(2y)$. This is an application of the classical quadratic threshold mechanism; a positive estimate requires actual edit resources beyond the endpoint and error allowance $B_y$.

With ${\mathcal{Q}}_s=Z(C_s{)}^s/C_s$, the exact same-host identity in FIB (234.10) shows that, for $N>5040$,

$$\boxed{G_{+}(z_{+})+G_{-}(z_{-})>\log {\mathcal{Q}}_s+\log N-s\log (e^{\gamma }\log \log N)}$$

is sufficient for Robin at that $N$. No superabundant property is needed. For a same-price optimizer $N=C_s$, however, $E_{+}=E_{-}=D_s(N)=0$; deleting a tied layer gives another zero-benefit optimizer. Consequently this certificate cannot exclude all low-loss candidates without additional same-host information. The FIB remainder, window and shared-divisor conditions have not been shown to force the displayed strict budget. The global RH objective remains unresolved.

Low increment loss supplies a smaller reconstruction height

The same certificate can also be used in the reverse direction to bound edit resources from an already known upper benefit. This application combines the classical capacity calculation above with the existing increment/benefit comparison in FIB (234.5)–(234.8); it makes no additional prime-distribution assumption.

Fix $h_0=1/3$. The common capacity bound is $A_{\pm }(h)\le (3y/2)h+B_y$. With $z_{\pm }=(E_{\pm }-B_y{)}_{+}$, layer-cake gives

$$D_s(N)\ge F_y(z_{+})+F_y(z_{-}),F_y(z)=\{\begin{array}{ll}{z}^2/(3y),& 0\le z\le y/2,\\ z/3-y/12,& z\ge y/2.\end{array}$$

Consequently $D_s(N)<y/12$ forces both $z_{\pm }<y/2$ and gives the finite resource bound

$$E_{+}+E_{-}\le 2B_y+\sqrt{6y{D}_s(N)}.$$

For a fixed $K>0$, put $\ell =\log y$, $R=y/{\ell }^2$ and take an arbitrary positive integer $d$ with the existing increment loss $J_s(d)\le KR$. Equation (234.5), with its nonnegative local correction, gives

$$D_s(d)\le J_s(d)+{\delta }_s\le KR+O(\log \ell )=o(y).$$

Since $B_y=O(y/{\ell }^2+\sqrt{y})=o(y/\ell )$, the preceding finite bound therefore yields, uniformly for this same low-loss set,

$$\sum _p|v_p(d)-v_p(C_s)|\log p=O_K(y/\ell ).$$

To transport this to the unchanged reference $n_y$ in FIB (230.4), note first that $y-1<x_s<y$: the lower inequality follows from $T(y-1)\le y\log (y-1)<s$. For primes above $H=\sqrt{s/\log 2}$, both references have exponent at most one, and they can differ only for a prime in $(x_s,y]$, an interval of length less than one. That resource is at most $\log y$. For $p\le H$, the geometric marginal bound gives $v_p(C_s)\log p\le \log s+O(1)$, and definition (230.4) gives $v_p(n_y)\log p=O(\log s)$. There are at most $H$ such primes. Thus

$$\sum _p|v_p(C_s)-v_p(n_y)|\log p=O(\sqrt{s}\log s+\log y)=o(y/\ell ).$$

The triangle inequality for the same prime-exponent resource now gives

$$\boxed{\mathcal{L}(d,n_y)=O_K(y/\log y)\text{uniformly when }{J}_s(d)\le Ky/(\log y{)}^2.}$$

This sharpens the $O_K(y/\sqrt{\log y})$ height input in FIB (230.7) by reuse of the benefit comparison and quantitative capacity. For a reduced ratio $d/n_y=\alpha /\beta$, it bounds both $\log \alpha$ and $\log \beta$. The whole-host reconstruction application retains its separate host-cofactor, integer-bound, unit, divisibility, window and source checks. An asymptotic height improvement is not an effective reconstruction threshold or a positive Robin margin. It does not prove that the actual host’s gcd resources satisfy the strict budget in the preceding section.

Charge the actual removals before bounding additions

The classical layer identity also permits a stronger pointwise use of the same gcd data. Put $g=\gcd (N,C_s)$ and $v=C_s/g$. The known factorization of $C_s$ fixes every removed layer, with exact cost

$$R_s(v)=\sum _{p\mid v}\sum _{j=a_p-v_p(v)+1}^{a_p}{c}_{p,j}=s\log \frac{Z(C_s)}{Z(C_s/v)}-\log v\ge 0.$$

Under the same finite capacity assumptions above, the already established addition bound and this exact removal cost give

$$\boxed{D_s(N)\ge R_s(v)+G_{+}(z_{+})\ge G_{-}(z_{-})+G_{+}(z_{+}).}$$

The two terms in the first lower bound pay disjoint parts of the existing layer decomposition: all actual removals and only the additions. The removal-side capacity bound gives $R_s(v)\ge G_{-}(z_{-})$, which proves the second comparison. Thus the sharper certificate is $R_s(v)+G_{+}(z_{+})>T_s(N)$ for the same actual host. Whenever $R_s(v)>G_{-}(z_{-})$, its lower bound increases by that difference. No factorization of $N/g$ is needed to evaluate either input.

This is a direct application of the classical layer decomposition and the existing capacity bounds, not a new benefit theorem or a Lean result. It does not assert that the improvement exceeds the actual budget on the FIB residual set. Tied removals retain their exact zero costs.

Exact zero-benefit exceptions use at most four branches

Proposition 4(c)–(d) and its proof, printed pp.70–71, already give the complete tied-optimizer alternatives. The proof uses the six-exponentials theorem, cited there through Lang, to exclude a common threshold for three distinct primes. It retains both exponent choices at each tied prime. This is a classical result to reuse, not a new tie theorem or a Lean result.

At the price $1/s$, keep the maximal optimizer $C_s$ and the layer notation above, and define

$${\mathcal{T}}_s=\{p\mid C_s:c_{p,v_p(C_s)}=0\}.$$

In these parameters the source’s alternatives state

$$|{\mathcal{T}}_s|\le 2,\{N\ge 1:D_s(N)=0\}=\left\{\frac{C_s}{{\prod }_{p\in S}p}:S\subseteq {\mathcal{T}}_s\right\}.$$

Thus the exact zero-benefit set has one, two or four members at every positive price. This conclusion concerns exact ties, not the number of near-zero layers or a uniform lower bound for the next positive cost.

For the actual host, $g=\gcd (N,C_s)$ fixes the denominator $v=C_s/g$ and therefore every removed layer. The zero-benefit alternative is precisely that $v$ is a product of a subset of ${\mathcal{T}}_s$ and $N=g$; it cannot become an unrestricted repetition of a free prime edit. One can retain the source’s at most four possibilities before computing the gcd, or use the actual gcd to determine the removals directly.

Each of these integers still requires the original FIB window, residue and qualifying-divisor checks. If an actual zero-benefit host has $T_s(N)<0$, its budget is already paid. If it has $T_s(N)\ge 0$, this classification provides no positive loss: a source-preserving exclusion or a favourable signed budget remains necessary. The fixed-price screening application keeps these ties when its endpoint budget is zero. Neither the finite branch list nor the gcd determines a uniform Robin margin.

Condition addition capacity on the actual removals

Keep the same $g=\gcd (N,C_s)$, $u=N/g$, $v=C_s/g$ and exact removal cost $R_s(v)$. Since $(u,v)=1$, no prime dividing $v$ can occur among the actual additions. The capacity calculation above can therefore exclude these primes before integration. Define, for $0\le h\le h_0\le 1/3$,

$${\Delta }_v(h)=\sum _{p\mid v}\sum _{\begin{array}{c}j>a_p\\ -c_{p,j}\le h\log p\end{array}}\log p,A_{+,v}(h)=A_{+}(h)-{\Delta }_v(h).$$

Both sums defining ${\Delta }_v$ are determined by the known factorization of $C_s$ and the actual gcd; factoring $u$ is unnecessary. Reuse $A_{+}(h)\le U_y(h):=yh/(1-h)+B_y$ at $s=y\log y$, $y\ge 6$. The addition-only layer-cake bound gives the paper-level refinement

$$\begin{array}{rl}{D}_s(N)& \ge R_s(v)+I_v(E_{+}),\\ I_v(E)& :={\int }_0^{h_0}(E-U_y(h)+{\Delta }_v(h){)}_{+}dh,\\ I_v(E_{+})& \ge G_{+}((E_{+}-B_y{)}_{+}),E_{+}=\log u.\end{array}$$

The subtraction is valid because $A_{+,v}$ is exactly the weak-layer capacity over primes eligible for the same host. It never removes a layer actually used by $u$. Exact removals and this addition integral still pay disjoint costs. This is an application of the existing finite layer-cake bound, not a new benefit theorem or a Lean-verified estimate.

The gain has a useful ceiling. Write

$${\tau }_p=\frac{-c_{p,a_p+1}}{\log p}>0(p\mid v).$$

The marginal separation ${\ell }_{p,j}/{\ell }_{p,j+1}>p$ proved above implies that every layer after $a_p+1$ has normalized cost greater than $1/2$. Thus only that first post-reference layer can contribute, and

$${\int }_0^{h_0}{\Delta }_v(h)dh=\sum _{p\mid v}(h_0-{\tau }_p{)}_{+}\log p.$$

For each nonzero summand, $a_p\ge 1$ and the same marginal separation gives

$$c_{p,a_p}>(p(1-{\tau }_p)-1)\log p\ge (1-2{\tau }_p)\log p.$$

Since $h_0-{\tau }_p\le h_0(1-2{\tau }_p)$ for $h_0\le 1/2$, and the last accepted layer at every $p\mid v$ is actually removed, it follows that

$$\boxed{0\le I_v(E_{+})-G_{+}((E_{+}-B_y{)}_{+})\le {\int }_0^{h_0}{\Delta }_v(h)dh\le h_0{R}_s(v).}$$

The first upper bound uses the one-Lipschitz property of the positive part. A tied removed layer has $c_{p,a_p}=0$ and forces ${\tau }_p>1-1/p\ge 1/2$, so it contributes nothing to this refinement. Excluding an exactly free removal cannot create an artificial positive gain.

Every contributing prime also satisfies $p^2\log p\le s/(1-h_0)\le 3s/2$, because $a_p+1\ge 2$. Consequently

$${\Delta }_v(h)\le \vartheta (P_2),I_v(E_{+})-G_{+}((E_{+}-B_y{)}_{+})\le h_0\vartheta (P_2)=O(\sqrt{y}).$$

This removes part of the higher-layer allowance for the actual gcd. The prime-error allowance $2{\eta }_y$ in $B_y$ is unaffected. The size bound does not give a positive lower bound for the gain or rule out payment of a smaller residual budget at a particular host. The remaining sufficient test is $R_s(v)+I_v(\log u)>T_s(N)$ for that same $N$; no uniform comparison on the original FIB residual sources has been established.

A full-modulus Jacobi test for actual addition cost

The character input here is classical quadratic reciprocity and Fibonacci modular periodicity, recorded with their primary locators in the Renault note. The application uses the benefit decomposition above; it supplies neither a new analytic character estimate nor a Lean-verified theorem.

Let $r>3$ be prime and $V=F_r$, with $F_0=0$, $F_1=1$. The Fibonacci pair modulo four has period six, and $r\equiv 1,5(\mathrm{mod}6)$, so $V\equiv 1(\mathrm{mod}4)$. Define the full-denominator Jacobi character

$${\chi }_V(a)=\left(\frac{a}{V}\right).$$

The denominator need not be prime or squarefree. For every odd prime $p$, quadratic reciprocity gives

$${\chi }_V(p)=\left(\frac{F_r}{p}\right),$$

including zero when $p\mid V$. The right-hand side reads $F_r\mathrm{mod}p$ from the second coordinate of $M^r\alpha$, where $M(a,b)=(b,a+b)$ and $\alpha =(1,0)$. This coordinate is different from the quantity readout $2a+3b$. The existing golden/Fibonacci modular pair in D5/S3/Arith/GoldenApparition.lean already supplies this interface; its private phi_pow_eq_fib_pair_mod is not a missing theorem to reprove. Neither the symbol evaluation nor the modular recurrence requires factorization of $V$.

For the small primes, the finite pair recurrences modulo eight, three and five have periods twelve, eight and twenty respectively. Together with the supplementary law at two and reciprocity they give the familiar specializations

$$\begin{array}{rl}{\chi }_V(2)& =\{\begin{array}{ll}1,& r\equiv 1,11(\mathrm{mod}12),\\ -1,& r\equiv 5,7(\mathrm{mod}12),\end{array}\\ {\chi }_V(3)& =\{\begin{array}{ll}1,& r\equiv 1,7(\mathrm{mod}8),\\ -1,& r\equiv 3,5(\mathrm{mod}8),\end{array}\\ {\chi }_V(5)& =\left(\frac{r}{5}\right)(r>5),{\chi }_{F_5}(5)=0.\end{array}$$

These symbol values are arithmetic observations, not the five containment labels [null,2,3,2 5,5]. For instance $r\equiv 1(\mathrm{mod}120)$ gives all three values $+1$; no fixed negative prime is forced by those three tests. This says nothing about signs throughout the growing price-prime prefix.

Now retain the same actual host $N>5040$ and require $N\equiv 1(\mathrm{mod}V)$. Use $C=C_s$, $g=\gcd (N,C)$, $u=N/g$ and $v=C/g$. Both $g$ and $u$ are units modulo $V$, since they divide $N$; this does not require $\gcd (C,V)=1$. Multiplicativity gives

$${\chi }_V(g){\chi }_V(u)=1,{\chi }_V(u)={\chi }_V(g).$$

In the branch ${\chi }_V(g)=-1$, some prime dividing the actual $u$ has negative character and odd exponent in $u$. It is eligible for additions: $p\nmid vV$. The negative endpoint also certifies nonprincipality, without a separate nonsquare theorem for $V$. A positive full Jacobi symbol at a composite denominator does not certify a square root; the branch ${\chi }_V(g)=1$ remains outside this particular test.

For eligible primes define the first-entry cost

$$b_s(p)=\log p-s{\ell }_{p,a_p+1}>0.$$

Maximality of $C$ includes every tied layer, so every post-reference cost is strictly positive. Its full addition block $W_p(k)=-{\sum }_{j=a_p+1}^{a_p+k}{c}_{p,j}$ is increasing for $k\ge 1$ and $W_p(1)=b_s(p)$. Thus the negative-endpoint branch has the paper-level bound

$$D_s(N)\ge R_s(v)+m_s(V,v),m_s(V,v):=\underset{\begin{array}{c}p\nmid vV\\ {\chi }_V(p)=-1\end{array}}{\min }{b}_s(p)>0.$$

The actual $u$ supplies an eligible negative prime. The minimum is attained because $C$ has finite support and $b_s(p)\to \infty$ as $p\to \infty$. This is a fixed-price positive bound, not a uniform gap along growing prices and moduli.

Use the inherited budget normalization

$$T_s(N)=\log {\mathcal{Q}}_s+\log N-s\log (e^{\gamma }\log \log N),{\mathcal{Q}}_s=Z(C_s{)}^s/C_s,$$

so $D_s(N)>T_s(N)$ is exactly strict Robin. Write $B=T_s(N)-R_s(v)$. If $B<0$, removals already pay the budget. For $B\ge 0$, define

$${\mathcal{P}}_B=\{p\text{ prime}:p\nmid vV,b_s(p)\le B\}.$$

A sufficient test for this same host is ${\chi }_V(g)=-1$ and ${\chi }_V(p)=1$ for every $p\in {\mathcal{P}}_B$. It gives $m_s(V,v)>B$ and hence $D_s(N)>T_s(N)$. Zeros cannot occur on this eligible set.

The test has an explicit finite cutoff. Put $P=\max \{s,e^{B+1}\}$. For $p>P$, one has $\log p>1$ and $p\log p>s$, so $a_p=0$ and

$$b_s(p)=\log p-s\log (1+1/p)\ge \log p-s/p>B.$$

Only primes $p\le P$ can belong to ${\mathcal{P}}_B$. This finiteness is not an efficiency bound at the actual source scale. Evaluating $g$, the exact removals and the symbol tests requires no factorization of $u$.

In the common capacity range $s=y\log y$, $y\ge 6$, $0<h_0\le 1/3$, the safe combination in the negative-endpoint branch is

$$D_s(N)\ge R_s(v)+\max \{I_v(\log u),m_s(V,v)\}.$$

Both lower bounds pay the same additions, so they cannot be summed without a further allocation to disjoint costs. The missing joint input is enough loss on each surviving actual source’s legal additions, compared with its own $B$. The one-bit sufficient test asks for a negative actual gcd and positive symbols at all eligible cheap primes. Neither modular periodicity nor the generic Pollack nonresidue bounds supplies that comparison at the source price. Failure of this sufficient test does not imply failure of Robin. The original window, qualifying low-loss divisor, cofactor and all-candidate coverage remain required; no uniform strict-budget supplier or proof of RH is supplied here.

The Fibonacci index prime limits the unrefined sign test

Use the actual family in FIB §§230–231, writing its multiplier as $k$ to distinguish it from the gcd:

$$V=F_r,\lceil V/10\rceil \le k\le \lfloor V/5\rfloor ,N=1+Vk>5040,A=1+V\lceil V/10\rceil ,y=\log A,s=y\log y,$$

where $r>5$ is prime. Keep the same $C_s,g,u,v$ and signed budget $B$. The ordinary Fibonacci bound $F_r<{\phi }^{r-1}$, together with $A<V^2$ and $2\log \phi <1$, gives $1<y<r$. Hence

$$s{\ell }_{r,1}=y\log y\log (1+1/r)<\frac{y\log y}{r}<\log r.$$

Thus $a_r=0$, $r\nmid C_s$ and $r\nmid v$, with strictness unaffected by tied layers. The already existing fibonacci_apparition_entry_point in D5/S3/Arith/GoldenApparition.lean supplies the classical congruence

$$F_r\equiv \left(\frac{5}{r}\right)(\mathrm{mod}r).$$

In particular $r\nmid V$. Reciprocity and the supplementary law at minus one give

$${\chi }_V(r)=\left(\frac{F_r}{r}\right)=\left(\frac{(5/r)}{r}\right)=-1⟺r\equiv 3,7(\mathrm{mod}20).$$

On these two classes the index prime is always eligible for the unrefined minimum, at exact cost

$$b_s(r)=\log r-s\log (1+1/r)>0,m_s(V,v)\le b_s(r).$$

Therefore $B\ge b_s(r)$ prevents the unrefined negative-gcd test $m_s(V,v)>B$ from succeeding. This is a limitation of that relaxation; it is not a Robin counterexample or an assertion that $r\mid u$. For the same actual host the congruence instead gives

$$r\mid u⟺r\mid N⟺k\equiv 1(\mathrm{mod}r).$$

The conditional threshold does not exhibit a surviving actual source with $B\ge b_s(r)$. The existing window budget gives $B\le T_s(N)\le T_s(X)$, where $X=1+V\lfloor V/5\rfloor$. If $T_s(X)<b_s(r)$, this index-prime obstruction never activates in that window. If $T_s(X)<0$, the window is already paid by the nonnegative benefit. Any claim of obstruction on the residual set requires an actual source there, retaining its low-loss divisor and cofactor.

Charge the actual index block before reading the remaining sign

On the same two negative-index classes, retain

$$e=v_r(N)=v_r(u),u_0=u/r^e,L_r=W_r(e)=e\log r-s\log Z(r^e),W_r(0)=0.$$

This is the exact cost of all additions at $r$, since $a_r=0$. The remaining factor has neither $r$ nor any prime dividing $vV$. For every $j\ge 1$, its required source resolution can be read as

$$r^j\mid N⟺k\equiv -F_r^{-1}(\mathrm{mod}{r}^j).$$

The inverse exists because $r\nmid V$; $F_r\mathrm{mod}{r}^j$ is the same modular $M^r\alpha$ readout used above. These are actual valuation probes, not a factorization of $V$ or an assumption about the unseen additions.

Multiplicativity transports the endpoint after charging that block:

$${\chi }_V(u_0)={\chi }_V(g)(-1{)}^e.$$

If this remaining sign is negative, the actual $u_0$ contains a distinct eligible negative-character prime. Define, only in that branch,

$$m_s^{[r]}(V,v)=\underset{\begin{array}{c}p\nmid vrV\\ {\chi }_V(p)=-1\end{array}}{\min }{b}_s(p)>0.$$

The actual factor ensures nonemptiness, and the same finite-tail argument as above ensures attainment. The paper-level certificate is

$$\{\begin{array}{ll}{D}_s(N)\ge R_s(v)+L_r+m_s^{[r]}(V,v),& {\chi }_V(g)(-1{)}^e=-1,\\ D_s(N)\ge R_s(v)+L_r,& {\chi }_V(g)(-1{)}^e=1.\end{array}$$

The index block and the distinct residual prime pay disjoint costs. If ${\chi }_V(g)=1$ and $e$ is odd, a second negative prime is forced, although the original negative-gcd test was inactive. If $e=0$ and ${\chi }_V(g)=-1$, the unused index prime disappears from the minimum. For an originally negative gcd, this bound is never weaker than $R_s(v)+m_s(V,v)$: when $e\ge 1$, $L_r\ge b_s(r)\ge m_s(V,v)$; when $e=0$, the new minimum ranges over a subset of the old one.

Put $B_r=T_s(N)-R_s(v)-L_r$. A negative $B_r$ is already paid. Otherwise, in the negative remaining-sign branch, $m_s^{[r]}(V,v)>B_r$ is sufficient. The same cutoff $P_r=\max \{s,e^{B_r+1}\}$ applies to this unrestricted cheap-prime test.

For pruning by the actual source, define a different minimum, only in the negative remaining-sign branch:

$$m_{s,\mathrm{act}}^{[r]}(N)=\underset{\begin{array}{c}p\nmid vrV,v_p(u_0)\mathrm{odd}\\ {\chi }_V(p)=-1\end{array}}{\min }{b}_s(p).$$

The negative endpoint supplies a nonempty subset of the finite actual prime support, so this minimum exists and is at least $m_s^{[r]}(V,v)$. The same disjoint-cost argument gives

$$D_s(N)\ge R_s(v)+L_r+m_{s,\mathrm{act}}^{[r]}(N).$$

With $B_r\ge 0$, actual probes may discard even $v_p(u_0)$, including zero. If every eligible $p\le P_r$ with $b_s(p)\le B_r$ and odd $v_p(u_0)$ has ${\chi }_V(p)=1$, the finite cutoff gives $m_{s,\mathrm{act}}^{[r]}(N)>B_r$ and hence strict Robin for this host. This need not imply $m_s^{[r]}(V,v)>B_r$: a cheap negative prime absent from $u_0$ can keep that unrestricted minimum small. No efficiency or uniform source correlation is asserted for the actual-support test.

When $y\ge 6$ and $0<h_0\le 1/3$, the previously established capacity bound still applies to all additions. It can be combined with the new bound by taking a maximum of their addition contributions. Summing that full capacity bound with $L_r+m_s^{[r]}$ would count the same additions twice. These are applications of the existing block decomposition, reciprocity and modular readout, with no new analytic theorem or Lean verification. A uniform estimate of the same actual remaining cost against $B_r$, with the original low-loss incidence and window retained, is still missing; existence of a second negative prime alone supplies neither its required cost nor a proof of RH.

Resolve the remaining arithmetic phase beyond its Jacobi sign

The additional character input is classical. Gao–Zhao, Value-distribution of quartic Hecke L-functions, arXiv:1809.09822v2, §2.2, PDF p.3, defines the Gaussian quartic power-residue symbol at an odd Gaussian prime and extends it multiplicatively to composite denominators. Only that definition is used here; the paper’s distribution theorem over square-free Gaussian parameters is not applied to this Fibonacci modulus. This is an application to the existing benefit certificate, not a new quartic-character theorem or a Lean result.

Retain the same actual family $V=F_r$, $N=1+Vk$, price $s=y\log y$ and maximal reference $C_s$ above, with $r>5$ prime. The classical Cassini identity gives $t=F_{r-1}$ with $t^2=-1(\mathrm{mod}V)$. Every prime $q\mid V$ is odd and $q\equiv 1(\mathrm{mod}4)$. Orient its quartic character by this root:

$${\psi }_q(n)=i^j⟺n^{(q-1)/4}\equiv t^j(\mathrm{mod}q),(n,q)=1,j\in \mathbb{Z}/4\mathbb{Z}.$$

In the Gaussian definition this uses the prime ideal $(q,i-t)$, whose residue field identifies $i$ with $t$. Retain all denominator exponents:

$${\psi }_V(n)=\prod _{q^E\parallel V}{\psi }_q(n{)}^E,(n,V)=1,{\psi }_V(n{)}^2={\chi }_V(n).$$

Neither primality nor square-freeness of $V$ is assumed. The resulting character can have order one, two or four, so four nonempty prime classes are not presumed. Identifying the residue root $t$ with complex $i$ does not assert ${\psi }_q(t)=i$. This arithmetic character is also different from the active composition rotation $C=MJ$ and from the existing scalar four-orbit analysis in §7 of the Li note. The factorization-free evaluation below uses a classical Gaussian quartic-Jacobi algorithm with this same root orientation.

Use the actual $g=\gcd (N,C_s)$, $u=N/g$, $v=C_s/g$ and $e=v_r(N)$, and set $u_0=u/r^e$. The existing $y<r$ argument gives $a_r=0$, and the existing Fibonacci entry-point congruence gives $r\nmid V$. These inputs hold on every prime-index residue class, not only the two classes with negative index-prime Jacobi sign. Charge the same exact block

$$L_r=W_r(e)=e\log r-s\log Z(r^e).$$

Since $g,u_0\mid N$ and $N\equiv 1(\mathrm{mod}V)$, both are units modulo $V$, without requiring $C_s$ to be a unit. Multiplicativity transports the remaining endpoint as

$${\psi }_V(u_0)={\psi }_V(g{)}^{-1}{\psi }_V(r{)}^{-e}.$$

This endpoint uses the same host and its actual index-prime valuation. When $u_0=1$, its phase is one and no remaining phase cost is forced.

For $j=1,2,3$, use the existing positive first-entry cost $b_s(p)$ and define

$${\beta }_j=\underset{\begin{array}{c}p\text{ prime},p\nmid vrV\\ {\psi }_V(p)=i^j\end{array}}{\min }{b}_s(p),$$

with value $+\infty$ for an empty class. Every nonempty class has a positive attained minimum: maximality of $C_s$ includes all ties, and $b_s(p)\to \infty$ at fixed price. This asserts no uniform gap as $s$ and $V$ grow.

The cheapest relaxed phase-word costs on $\mathbb{Z}/4\mathbb{Z}$ are

$$\begin{array}{rl}{c}_0& =0,\\ c_1& =\min \{{\beta }_1,{\beta }_2+{\beta }_3,3{\beta }_3\},\\ c_2& =\min \{{\beta }_2,2{\beta }_1,2{\beta }_3\},\\ c_3& =\min \{{\beta }_3,{\beta }_1+{\beta }_2,3{\beta }_1\}.\end{array}$$

To use this relaxation, retain the actual prime-power word of $u_0$. The increasing post-reference marginal costs give $W_p(k)\ge k{b}_s(p)$, so repeated occurrences of one prime are charged. Replace each actual nonzero-phase occurrence by its class minimum. Phase-zero edges and positive-cost loops can then be deleted. A shortest nonzero-endpoint path visits at most the four states; its undominated increment words are $1,23,333$ for endpoint one, $2,11,33$ for endpoint two, and $3,12,111$ for endpoint three. This gives, for the actual endpoint ${\psi }_V(u_0)=i^j$, the paper-level application

$$\boxed{D_s(N)\ge R_s(v)+L_r+c_j.}$$

An actual endpoint guarantees a finite route even if some classes are empty. The relaxed minimizing word need not realize its class minima simultaneously, the real size window or the low-loss incidence; these facts are not required for a lower bound. They would be required to claim an actual realizing candidate.

For odd $j$, the unrefined remaining Jacobi minimum is $m_s^{[r]}(V,v)=\min \{{\beta }_1,{\beta }_3\}$ and $c_j$ is at least this minimum. If ${\beta }_3<{\beta }_1$, then $c_1$ is strictly larger; if ${\beta }_1<{\beta }_3$, then $c_3$ is strictly larger. The relevant inequalities allow an empty opposite class. A reachable phase two has $c_2>0$, although its remaining Jacobi sign is positive. These are conditional improvements over the sign-only bound. There is no universal ordering against $m_{s,\mathrm{act}}^{[r]}(N)$: cheap eligible primes absent from the host can reduce $c_j$, whereas a required repeated phase can force more cost than one actual negative-prime minimum. Complete actual valuations already determine these phases.

With $y\ge 6$ and $0<h_0\le 1/3$, the safe capacity combination is

$$D_s(N)\ge R_s(v)+\max \{I_v(\log u),L_r+c_j\}.$$

In the negative remaining-sign branch one may also include $L_r+m_{s,\mathrm{act}}^{[r]}(N)$ inside this maximum. These contributions bound the same additions; summing the full capacity contribution with $L_r+c_j$ would pay them twice.

For the same host put $B_r=T_s(N)-R_s(v)-L_r$. The condition $c_j>B_r$ is sufficient to pay its strict budget. No such comparison on every qualifying host has been established, and failure of this sufficient condition does not disprove Robin. Phase zero, $u_0=1$ and all zero-cost tied removals remain in scope; the actual gcd fixes each removed layer once, rather than licensing a repeatable free phase loop. The original window and the complete low-loss divisor contribution in FIB §233.5 still require a joint estimate. This includes $h=1$, restoration-only and small-prime cofactors with their actual merged valuations. This character refinement alone supplies no uniform strict budget or RH proof.

Reuse the Gaussian quartic-Jacobi evaluator

The algorithmic input is already available in Damgård–Frandsen, Efficient algorithms for gcd and cubic residuosity in the ring of Eisenstein integers, BRICS RS-03-8, §5, printed pp.9–10, PDF pp.11–12. Despite its title, that section treats Gaussian gcd and composite quartic symbols, with quadratic bit complexity for its specified approximate-norm method. Bach–Sandlund, On Euclidean Methods for Cubic and Quartic Jacobi Symbols, arXiv:1807.07719v1, §7, pp.13–14, distinguishes the naive exact-norm Euclidean implementation, which has cubic worst-case schoolbook complexity, from the improved quadratic quotient treatment. These algorithms are reused; no arithmetic algorithm or new analytic estimate is proposed here.

For the same $V=F_r$ and $t=F_{r-1}$, form the Gaussian ideal

$$I=(V,i-t)=\ker (\mathbb{Z}[i]⟶\mathbb{Z}/V\mathbb{Z},a+bi⟼a+bt).$$

The map is well-defined because $t^2=-1(\mathrm{mod}V)$, and it is surjective. Thus $\mathbb{Z}[i]/I\simeq \mathbb{Z}/V\mathbb{Z}$ and $I$ has norm $V$. Since $\mathbb{Z}[i]$ is Euclidean, a denominator is

$$\eta =\mathrm{primary}\underset{\mathbb{Z}[i]}{\gcd }(V,t-i),\mathrm{Nm}(\eta )=V.$$

Here primary means $\eta =a+bi$ with $b$ even and $a+b\equiv 1(\mathrm{mod}4)$. For each conceptual factor $q^E\parallel V$, only the selected prime ideal $(q,i-t)$ divides $t-i$; the conjugate prime does not, since $2t$ is nonzero modulo $q$. Taking the gcd with $V$ retains exactly $E$ copies of the selected prime. Consequently the previously defined full-denominator character is precisely

$$\boxed{{\psi }_V(n)={\left(\frac{n}{\eta }\right)}_4,(n,V)=1.}$$

The factorization in this explanation verifies the input translation; the gcd and quartic-Jacobi algorithms do not require it. They must not replace the denominator by its radical or its primitive-character part. Two useful input checks are $a^2+b^2=V$ and $a+bt\equiv 0(\mathrm{mod}V)$.

Multiplying $\eta$ by a Gaussian unit preserves its ideal and its symbol; conjugating it reverses the chosen orientation. Using the rational Gaussian integer $V$ as denominator instead includes both conjugate factors and gives

$${\left(\frac{n}{V}\right)}_{4,\mathbb{Z}[i]}={\psi }_V(n)\overline{{\psi }_V(n)}=1(n,V)=1,$$

so that substitution would erase the required phase.

Numerator normalization has a different rule. The supplementary laws in the same Gao–Zhao §2.2 source give, for primary $\eta =a+bi$,

$${\left(\frac{i}{\eta }\right)}_4=i^{(1-a)/2},{\left(\frac{1+i}{\eta }\right)}_4=i^{(a-b-b^2-1)/4}.$$

If $z=i^j(1+i{)}^h{z}_0$ with $z_0$ primary, these removed numerator factors contribute

$${\left(\frac{z}{\eta }\right)}_4=i^{j(1-a)/2+h(a-b-b^2-1)/4}{\left(\frac{z_0}{\eta }\right)}_4.$$

The full published algorithm already tracks them. A wrapper that makes the numerator primary and discards the factors changes the answer, including at the cheap prime two. Since $2=-i(1+i{)}^2$, one obtains

$$\boxed{{\psi }_V(2)=i^{-b/2}.}$$

With $L=\lceil {\log }_{2}V\rceil$, the cited quadratic algorithms give $O(L^2)$ bit operations for denominator construction and for each query after reducing a rational numerator modulo $V$. Reading and reducing an unreduced input is a separate cost. This bound is in the bit size of $V$, not in the bit size of the Fibonacci index $r$, and does not apply to arbitrary exact-norm Euclidean code.

For the actual endpoint, cache $\eta$ and evaluate $(g/\eta {)}_4$ and $(r/\eta {)}_4$, then use

$${\psi }_V(u_0)={\left(\frac{g}{\eta }\right)}_4^{-1}{\left(\frac{r}{\eta }\right)}_4^{-e}.$$

This step needs neither factorization of $V$ nor of $u_0$; the actual valuation $e$ is still an input, with only $e\mathrm{mod}4$ needed for this phase. The same evaluator labels eligible cheap primes for the existing ${\beta }_j$ minima. It supplies their phase labels, not their minimum cost, their occurrence in the same actual host, or $c_j>B_r$. The full weighted Robin budget in FIB §233.5 remains a separate unresolved estimate.

The half-index Fibonacci pair already gives the denominator

For this particular modulus, even the generic Gaussian gcd preprocessing can be omitted. The Fibonacci doubling and Cassini identities are classical; the pinned Mathlib sources already contain Nat.fib_two_mul_add_one and Nat.fib_two_mul and Int.fib_succ_mul_fib_pred_sub_fib_sq. Their application here is a denominator recipe, not a new Fibonacci identity or a compiled Lean bridge.

Write $r=2m+1$, $A=F_{m+1}$ and $B=F_m$. The same identities give

$$V=A^2+B^2,t=B(2A-B),A^2-AB-B^2=(-1{)}^m,$$