Erdős 699: weighted certificates make index 21 effective

P699 research · Erdős problem 699 · Small-index bounds

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Completed follow-up. The cutoff-21 proof completely excludes index $21$ and every numerically bounded case in this note. The weighted inequalities below supply their analytic tails.

Index 21 has an effective last row. A weighted polynomial gives a positive exponent even though the unweighted small-prime estimate misses the required exponent. The same construction substantially reduces the algebraic terms in several existing stopping bounds. Coinciding maximal positions of different small primes are included in the proof.

The exact checker and recorded output contain the exponent optimizations and integer comparisons. They enumerate finite sets of exponent vectors and subsets of small primes, not binomial rows or columns.

1. Weighted divisibility

Let $S_i=\{p:p<i\}$, $s=|S_i|$, and choose a position $r_p\in\{0,\ldots,i-1\}$ maximizing $v_p(n-r_p)$. Write

$$P_p=p^{v_p(n-r_p)},\qquad u_p=(n-r_p)/P_p, \qquad h=n-i+1.$$

For an integer $k\le i$ with $2k\ge i-1$, put

$$F_k(X,Y)=\prod_{a=0}^{k-1}((X-a)(Y-a))^{k-a},\quad D=k(k+1),\quad m_t=2k-t,$$

and $M=\sum_{t=0}^{i-1}m_t$, $L=\sum_{t=0}^{i-1}t m_t$. The position-sensitive construction proves the required vanishing at the Kummer points, and $0<F_k(j,n-j)\le n^D/2^D$ on every admissible pair.

Here we retain the exact constants away from the selected maxima. Define

$$A_p=\max_{0\le r<i}\sum_{\substack{0\le t<i\\t\ne r}} m_t v_p(|t-r|),\qquad C_* =\prod_{p<i}p^{A_p}, \qquad H=\begin{cases}i&i\text{ prime},\\1&i\text{ composite}.\end{cases}$$

The maximality of $r_p$ implies $v_p(n-t)\le v_p(t-r_p)$ for $t\ne r_p$. Also $\prod_t(n-t)^{m_t}\ge n^M(1-L/n)$. Thus badness gives

$$\prod_{p<i}P_p^{m_{r_p}} \ge\frac{2^D(1-L/n)n^{M-D}}{H^{2k}C_*}. \tag{1}$$

The factor $H^{2k}$ covers the rough prime in $i!$ at prime indices. Formula (1) is valid whether or not the selected positions coincide.

2. All position assignments, including coincidences

The source estimates supply maximum constraints

$$\max\left\{\frac{\log u_p}{\log h}, \frac{\log u_q}{\log h}\right\}\ge\lambda_{pq}.$$

Select a witnessing endpoint of each constraint. At a vertex, take the largest weight assigned to it. Removing componentwise dominated vectors leaves a finite list of $v=(v_p)$, one of which satisfies

$$u_p\ge h^{v_p},\qquad P_p\le n^{1-v_p}(n/h)^{v_p}. \tag{2}$$

Every vector used here has $0\le v_p<1/2$.

Group the primes having the same selected position. A singleton group $\{p\}$ has exponent capacity $c=1-v_p$. A group with at least two primes has capacity $c=1$, because the product of its distinct prime powers divides the single integer $n-r$ and is at most $n$. Applying (2) only at singleton groups therefore gives an upper bound for the left side of (1) of the form

$$n^{\sum_r(2k-r)c_r}(n/h)^B,\qquad 0\le B\le2ks. \tag{3}$$

For fixed capacities, their largest possible weighted sum occurs at the positions $0,1,\ldots,g-1$, with the capacities in decreasing order. This is just rearrangement, since every capacity is positive and the weights $2k-r$ decrease with $r$.

It remains to cover all ways of forming groups. Choose the subset $J$ of primes at singleton positions. If $b=s-|J|$ primes remain, then $b\ne1$; when $b\ge2$, there are at most $\lfloor b/2\rfloor$ other occupied positions. Each has capacity one. The worst possibility has exactly that many: inserting another capacity one at the start increases the weighted sum by at least $2k-g\ge0$. Hence it suffices to consider the finite capacity list

$$\underbrace{1,\ldots,1}_{\lfloor(s-|J|)/2\rfloor}, \quad (1-v_p)_{p\in J}, \tag{4}$$

sorted in decreasing order. This covers arbitrary collisions using only $2^s$ subsets, rather than enumerating positions or set partitions.

Let $c_0\ge\cdots\ge c_{g-1}$ denote a list (4). The gain to check is

$$G(v,J)=M-D-\sum_{r=0}^{g-1}(2k-r)c_r. \tag{5}$$

If every gain is at least $\eta>0$, combine (1)--(3). For

$$n\ge N:=1000\bigl(L+2ks(i-1)\bigr),$$

the elementary product inequality gives

$$(1-L/n)(1-(i-1)/n)^{2ks}\ge999/1000.$$

Consequently every bad row above the source thresholds satisfies

$$\boxed{\displaystyle n^\eta\le\frac{1000}{999}\,\frac{H^{2k}C_*}{2^D}.} \tag{6}$$

All quantities on the right are explicit rational numbers.

3. The new index and smaller algebraic bounds

For index 21, take $k=14$ and the published BFT Corollary 2.3 table, augmented by

$$(2,17):\ .3560,\qquad (17,19):\ .2176.$$

Both additional exponents are certified applications of BFT Theorem 2.2; see the new prime-pair estimates. The finite calculation above has minimum gain

$$\eta_{21}=\frac{7683}{5000}>0.$$

Its worst case has all eight positions distinct, and the cofactor vector in prime order $2,3,5,7,11,13,17,19$ is

$$(.3560,.3453,.2667,0,0,.1243,0,.2176).$$

Equation (6) supplies the algebraic term $10^{232}$. Hence

$$n<\max\{10^{232},X_{21}+20\}, \tag{7}$$

where $X_{21}$ is the maximum of the finitely many effective source thresholds for these pairs. This proves effectivity independently of evaluating those thresholds numerically.

For comparison, retaining the same sources as the previous effective bounds gives:

The first four use only the published Corollary 2.3 table. The last uses that table and the already established pairs $(17,19):.21$ and $(11,23):.145$. These improved algebraic terms do not discard the source thresholds: the complete bounds remain their maxima with the applicable source threshold and its row shift.

4. Fully numerical versions

Smaller, explicitly certified prime-pair exponents can give stronger complete bounds because their source thresholds are much smaller. The numerical source calculations and explicit new pairs supply the required thresholds. With the original, fully explicit BFT Theorem 2.1 table as the starting point, the following choices give:

At indices 19, 22, 24, and 25, the source checker verifies the row shifts up to 33 directly, so no extra power of ten is required. At index 16, the source threshold is below $10^{163\,784}$ and the displayed extra power of ten absorbs the row shift.

At index 21, use the numerical refinements of the old table, including $(5,7):.2587$, $(7,13):.1184$, and $(3,11):.3432$, together with $(2,17):.3379$ and $(17,19):.1925$. The same $F_{14}$ calculation has gain $1711/10000$ and algebraic term $10^{2075}$. Every source threshold is below $10^{45\,411\,830}$, so allowing the row shift gives the fully numerical bound

$$\boxed{n<10^{45\,411\,831}\qquad(i=21).} \tag{8}$$

These are stopping bounds. No search through all rows below them is claimed in this note.