Erdős 699: a complete effective cutoff certificate at index 18

P699 research · Weighted certificates · Asymmetric cofactor bounds

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Subsequent result. The complete index-18 proof instantiates a numerical cutoff and excludes its entire remaining interval. There are no bad triples with $i=18$. The earlier effective reduction is retained below.

Every bad triple with $i=18$ lies below an effectively computable last row. This note supplies the complete finite arithmetic certificate and includes the source thresholds in the final bound.

Let $X_{18}$ be the maximum of the effective bounded-gap thresholds, with gap $D=17$, for the symmetric constraints used below. The thresholds come from BFT Corollary 2.3 and the verified new prime-pair estimates. Then an exclusive bound is

$$\boxed{n<\max\{X_{18},10^{2\,405\,782}\}+17.}\tag{1}$$

The number $X_{18}$ is effectively computable but has not been evaluated numerically here. Thus (1) is an effective bound, not a fully numerical bound or an assertion that the exceptional set is empty.

Source constraints and the finite certificate

Put $S=\{2,3,5,7,11,13,17\}$. For each $p\in S$, choose $r_p\in\{0,\ldots,17\}$ maximizing $v_p(n-r_p)$ and write

$$n-r_p=P_pu_p,\qquad P_p=p^{v_p(n-r_p)},\qquad h=n-17.$$

The inputs associated with any two primes differ by at most $17$. Use every symmetric edge from BFT Corollary 2.3 and the new-pair table whose two primes belong to $S$. Add exactly two asymmetric constraints:

$$\begin{aligned} u_5>h^{3613/10000}&\quad\text{or}\quad u_7>h^{2277/10000},\\ u_7>h^{987/10000}&\quad\text{or}\quad u_{13}>h^{2080/10000}. \end{aligned}\tag{2}$$

The asymmetric source checker proves (2) once all selected inputs exceed, respectively,

$$10^{2\,405\,782},\qquad10^{484\,367}.\tag{3}$$

These are valid for gaps through $100$, so they apply here. At the row level it is enough to add $17$ to each threshold in (3).

Write $z_p=\log u_p/\log h$. Each symmetric or asymmetric constraint has the form $z_p\ge a$ or $z_q\ge b$. Choosing a witnessing endpoint of each constraint and discarding componentwise larger vectors produces 65 minimal vectors. Every admissible real vector $z$ dominates one of them: when another constraint is added, raise an endpoint that witnesses that constraint for $z$; removing a dominated vector preserves this property.

The exact checker constructs these vectors with integer weights scaled by $10^4$. Its recorded output contains all 65 vectors, every source edge, the worst collision capacity list, and the arithmetic constants below. It also recomputes the two interval proofs giving (3). No binomial rows are searched.

The polynomial and all collisions of maximal positions

Take

$$F(X,Y)=\prod_{a=0}^{11}((X-a)(Y-a))^{12-a},\qquad D=156,\qquad m_t=24-t\quad(0\le t<18).$$

At a Kummer point $(r,s)$ with $r+s=t$, the vanishing order of $F$ is at least $m_t$. Also $0<F(j,n-j)\le n^{156}/2^{156}$. The weighted divisibility argument therefore gives

$$\prod_{p\in S}P_p^{m_{r_p}} \ge\frac{2^{156}(1-1887/n)n^{123}}{C_*},\tag{4}$$

where the exact contribution away from each selected maximum is bounded by

$$\begin{aligned} C_*&=\prod_{p\in S}p^{A_p},\\ A_p&=\max_{0\le r<18}\sum_{\substack{0\le t<18\\t\ne r}} (24-t)v_p(|t-r|),\\ C_*&=2^{272}3^{108}5^{57}7^{41}11^{24}13^{24}17^{24}. \end{aligned}\tag{5}$$

Indeed $v_p(n-t)\le v_p(t-r_p)$ away from a maximal position. The remaining constants in (4) are $\sum m_t=279$ and $\sum t m_t=1887$.

For a minimal vector $v$ dominated by $z$, a singleton position occupied by prime $p$ has exponent capacity $1-v_p$. A position shared by two or more primes has capacity $1$, since the product of those prime powers divides one integer $n-r$. For each possible subset $J$ of singleton primes, the worst list of capacities is

$$\underbrace{1,\ldots,1}_{\lfloor(7-|J|)/2\rfloor}, \qquad (1-v_p)_{p\in J},\tag{6}$$

sorted in decreasing order; omit the impossible case $7-|J|=1$. Rearrangement puts these capacities at positions $0,1,\ldots$ in the worst case. These lists include arbitrary coincidences of maximal positions, as proved in the weighted-certificate note.

The checker evaluates 5707 vector-and-capacity cases across the 65 vectors and proves

$$123-\sum_{r=0}^{g-1}(24-r)c_r\ge\frac{186}{625}.\tag{7}$$

Equality occurs at

$$v=(.3560,.3453,.2091,.2667,0,.0729,0),$$

in prime order, with capacities

$$(1,1,.9271,.7909,.7333,.6547,.6440).$$

The replacement $P_p\le n^{1-v_p}(n/h)^{v_p}$ introduces a factor at most $(n/h)^{168}$. For

$$n\ge1000(1887+168\cdot17)=4\,743\,000,$$

we have $(1-1887/n)(h/n)^{168}\ge999/1000$. Combining (4)--(7) therefore gives

$$n^{186/625}\le\frac{1000}{999}\frac{C_*}{2^{156}}.\tag{8}$$

The checker clears the rational exponent and compares integer powers to prove that (8) requires $n<10^{814}$.

The full stopping bound

All parts of the argument apply when

$$n\ge\max\{10^{814},X_{18}+17, 10^{2\,405\,782}+17,10^{484\,367}+17\}.$$

Then (8) contradicts $n\ge10^{814}$. Thus every bad row is below that maximum, which equals the bound in (1). The large asymmetric thresholds are essential terms in this statement; the algebraic term $10^{814}$ alone is not a complete stopping bound.