Erdős 699: an effective bound at index 11

P699 research · Erdős problem 699 · Position-sensitive certificates

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Complete exclusion. The index-eleven proof excludes every bad triple with $i=11$. The bounds below remain intermediate inputs.

Numerical alternative. The explicit source-threshold calculation now also proves the unconditional bound $n<10^{3\,639\,160\,902}$. The argument below keeps the original stronger cofactor exponents and improves their algebraic term to $10^{484}$.

There is an effectively computable last row for bad triples with $i=11$. The essential estimate has a positive exponent $7/100$. More precisely, let $X$ be the maximum of the six effective constants $x_0(10)$ supplied by Bennett–Filaseta–Trifonov, Corollary 2.3, for pairs from $\{2,3,5,7\}$. Every bad triple with $i=11$ satisfies

$$n<\max\{10^{484},X+10\}. \tag{1}$$

The algebraic term $10^{484}$ is certified explicitly. The six source constants comprising $X$ are effectively computable, but have not been numerically evaluated here. Thus (1) is an effective bound, not a claim that $10^{484}$ alone bounds every bad row. The sharper constant calculation below improves the original algebraic term $10^{1525}$.

The original threshold $p\ge11$ is used throughout. In particular, 11 is not a small prime, and its contribution to $11!$ must be retained.

The prime-pair input

For each $p\in S=\{2,3,5,7\}$, choose $r_p\in\{0,\ldots,10\}$ maximizing $v_p(n-r_p)$, and write

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

The BFT paper, Corollary 2.3, proves that for sufficiently large, effectively bounded $n-r_p$,

$$\max(u_p,u_q)>(n-r_p)^{\lambda_{pq}},$$

because $|(n-r_p)-(n-r_q)|\le10$. The required values are

Assume $n\ge\max\{20,X+10\}$ and set

$$e_p=\frac{\log u_p}{\log(n/2)}.$$

Then $\max(e_p,e_q)>\lambda_{pq}$ for every pair. For each of the six pairs, select an endpoint witnessing its inequality, and assign to each vertex the largest edge weight assigned to it. This gives a vector $v=(v_2,v_3,v_5,v_7)$ with $0\le v_p\le e_p$.

Enumerating the $2^6=64$ endpoint choices and removing vectors that are componentwise larger than another leaves exactly the following eight. The last column is an exact rational calculation used below.

Thus one of these eight vectors is componentwise at most $e$. In particular,

$$P_p=\frac{n-r_p}{u_p} \le 2^{v_p}n^{1-v_p}. \tag{2}$$

The input is a maximum constraint, not merely $e_p+e_q>\lambda_{pq}$. Replacing it by the latter loses the saving needed here.

The weighted polynomial

Use the polynomial from the position-sensitive construction with $k=7$:

$$F(X,Y)=\prod_{a=0}^6((X-a)(Y-a))^{7-a}.$$

It has degree $D=56$, and at every Kummer point on $x+y=t\le10$ its multiplicity is at least

$$m_t=14-t.$$

Here $M=\sum m_t=99$ and $L=\sum t m_t=385$. The weighted valuation estimate proved in the position-sensitive note, equation (7), therefore gives, for $n\ge770$,

$$\prod_{p\in S}P_p^{14-r_p} \ge\frac{2^{55}n^{43}}{(11!)^{14}}. \tag{3}$$

This uses $[11!]_{\{p\ge11\}}=11$, explaining the $11!$ in the denominator.

First suppose the four positions $r_p$ are distinct. Combine (2) and (3). The resulting exponent of $n$ is

$$G(v,r) =43-\sum_{p\in S}(14-r_p)(1-v_p) =14\sum_pv_p-13+\sum_pr_p(1-v_p). \tag{4}$$

Every coefficient $1-v_p$ is positive. Hence the minimum over distinct nonnegative integer positions occurs at $\{0,1,2,3\}$. Rearrangement assigns the largest $v_p$ to the largest position. If $v_{(0)}\le\cdots\le v_{(3)}$ are the sorted coordinates, this minimum is

$$-7+\sum_{a=0}^3(14-a)v_{(a)}. \tag{5}$$

The eight exact evaluations in the table show

$$G(v,r)\ge\frac7{100}>0 \tag{6}$$

for every assignment of the four primes to distinct positions, without enumerating the 7920 assignments individually.

As $v_p<1$ and $14-r_p\le14$, we can bound the harmless power of two in (2) by $2^{56}$. Equations (3)--(6) yield

$$n^{7/100}\le 2(11!)^{14}. \tag{7}$$

The exact integer comparison

$$10^{1525\cdot7}>\bigl(2(11!)^{14}\bigr)^{100}$$

then excludes $n\ge10^{1525}$ once the BFT thresholds have been reached.

If the selected positions are not distinct, at most three numerator factors contain all four selected maxima. The degree-reduction bound in the effectivity note gives $U_{11}(n)\le n^3$. The existing uniform certificate has $V_{11}(n)^{11}<n^{84}/4^{28}$, so for $n\ge110$,

$$n^4<\frac{(2\cdot11!)^{11}}{4^{28}}.$$

In particular $n<10^{22}$. This also lies below the algebraic cutoff in (1), completing the effective reduction.

The exact checker verifies the finite endpoint optimization, all eight gains, and the displayed integer comparisons. It does not enumerate rows or columns.

Sharper constants: from $10^{1525}$ to $10^{484}$

Keep the exact contributions away from the selected maximum. Define

$$A_p(r)=\sum_{\substack{0\le t\le10\\t\ne r}} (14-t)v_p(|t-r|).$$

The ultrametric estimate gives the following stronger version of (3):

$$\prod_{p\in S}P_p^{14-r_p} \ge \frac{2^{56}n^{43}\prod_{t=0}^{10}(1-t/n)^{14-t}} {11^{14}\prod_{p\in S}p^{A_p(r_p)}}. \tag{8}$$

Here $11^{14}$ bounds the removed rough denominator, and each other denominator factor is bounded separately at its actual selected position. Normalize the cofactors by $n-10$ instead of $n/2$. The same eight vectors then give

$$P_p\le n^{1-v_p}(1-10/n)^{-v_p}.$$

Since $\sum(14-r_p)v_p\le56$, combining this with (8) shows

$$n^{G(v,r)} \le \frac{1000}{999}\, \frac{11^{14}\prod_{p\in S}p^{A_p(r_p)}}{2^{56}} \qquad(n\ge10^6). \tag{9}$$

Indeed, the two factors moved to the denominator have product at least $(1-385/n)(1-560/n)>999/1000$. This follows from the elementary product inequality and Bernoulli's inequality; the displayed numerical comparison holds already at $n=10^6$.

There are $11\cdot10\cdot9\cdot8=7920$ assignments of distinct positions and eight cofactor vectors. Evaluating (9) for these 63,360 proof patterns, with outward-rounded logarithms, gives

$$\log_{10}n<483.776812.$$

The largest ratio occurs at

$$(r_2,r_3,r_5,r_7)=(3,0,2,1),\qquad v=(.2921,0,.2667,.0505),$$

where $G=7/100$ and $(A_2,A_3,A_5,A_7)=(41,29,7,6)$. Thus the right side of (9) is $\frac{1000}{999}3^{29}5^7 7^6 11^{14}/2^{15}$ in this worst pattern. The collision bound $10^{22}$ and the starting threshold $10^6$ are both smaller than $10^{484}$, proving (1). The refined checker and its certificate retain every comparison needed for this bound. This calculation checks the finitely many proof patterns, not binomial rows.

Why the same calculation stops at index 13

For $i=13$, the graph on $\{2,3,5,7,11\}$ has 15 componentwise minimal vectors. With the analogous polynomial $F_8$, fourteen give a positive gain after the same rearrangement argument. The sole remaining vector is

$$(v_2,v_3,v_5,v_7,v_{11}) =(0.2757,0.3453,0.2667,0,0),$$

whose minimum gain is $-1077/2000$. Thus this calculation does not make index 13 effective. The BFT table has no entry for the pair $(7,11)$; after Corollary 2.3 and again in Section 8, the authors explicitly identify that pair as one for which their construction supplies no positive bound. The index-11 conclusion does not depend on filling that gap.