Erdős 699: certificates adapted to the small-prime positions

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This is a retained earlier research note. The current reading guide gives the latest status and distinguishes completed proofs from remaining cases.

P699 research · Erdős problem 699 · Effectivity reduction

The uniform certificate discards which numerator factor contains the largest power of each small prime. Retaining these positions gives effective bounds for additional families. It does not give a row bound for an entire new index.

At $i=27,30,33$, put $s=\pi(i-1)$, so $i=3s$ and respectively $s=9,10,11$. For a bad triple beyond an effectively computable bound, the selected positions of the $s$ small primes must be distinct. If their set is $T$, then

$$\sum_{t\in T}t\le\frac{s(s+1)}2-1, \qquad \max T\le2s-2. \tag{1}$$

There are respectively 67, 97, and 139 sets $T$ satisfying (1), before assigning the individual small primes to these positions. This is a finite list of position patterns, each still allowing unbounded rows. It is not a finite exceptional set of triples.

Superseded. The prime-pair estimates exclude all three indices $27,30,33$ completely. The weighted construction below also supplies the effective index-eleven proof.

A weighted polynomial

As usual, $c=n-j$, $i<j\le n/2$, and badness means that no prime $p\ge i$ divides both binomial coefficients. For each prime $p<i$, select $r_p\in\{0,\ldots,i-1\}$ maximizing $v_p(n-r_p)$, and put

$$P_p=p^{v_p(n-r_p)},\qquad T=\{r_p:p<i\}.$$

For an integer $1\le k\le i$, take

$$F_k(X,Y)=\prod_{a=0}^{k-1} ((X-a)(Y-a))^{k-a},\qquad D=k(k+1). \tag{2}$$

At each Kummer point $(x,y)$ with $x+y=t<i$, its multiplicity is

$$(k-x)_++(k-y)_+\ge m_t:=(2k-t)_+.$$

The polynomial is positive at $(j,c)$, and pairing the factors gives

$$0<F_k(j,c)\le n^D/2^D. \tag{3}$$

Let $Q_t$ be the part of $n-t$ supported on primes at least $i$, with the corresponding denominator factor from $i!$ removed. Thus $\prod_tQ_t=V_i(n)$. Under badness, a prime-power component of $Q_t$ forces $(j,c)$ to reduce to one of the Kummer points on $x+y=t$. Taylor expansion of the integer polynomial (2) therefore gives

$$\prod_{t=0}^{i-1}Q_t^{m_t}\mid F_k(j,c). \tag{4}$$

This remains valid at prime indices. If $i$ is prime, the factor $i!$ removes one power of $i$ from the unique numerator factor it divides; every power remaining in $Q_t$ still comes from an active prime.

Put

$$H=[i!]_{\{p:p\ge i\}} =\begin{cases}1&i\text{ composite},\\i&i\text{ prime},\end{cases} \quad M=\sum_t m_t,\quad L=\sum_t t m_t, \quad C=2^{1-D}\bigl(H(i-1)!\bigr)^{2k}. \tag{5}$$

For each $p<i$ and $t\ne r_p$, maximality gives $v_p(n-t)\le v_p(t-r_p)$. Since $m_t\le2k$, the factors away from the selected position contribute at most

$$\prod_{p<i}p^{\sum_{t\ne r_p}m_t v_p(n-t)} \le ((i-1)!)^{2k}. \tag{6}$$

The removed rough denominator contributes at most $H^{2k}$. For $n\ge2L$, the product inequality

$$\prod_t(1-t/n)^{m_t}\ge1-L/n\ge\tfrac12$$

and (3)--(6) now imply

$$\boxed{\displaystyle \prod_{p<i}P_p^{m_{r_p}}\ge\frac{n^{M-D}}{C}.} \tag{7}$$

In particular, if the selected positions are distinct and

$$E=\sum_{t\notin T}m_t-D>0,$$

then $P_p\le n$ gives the explicit bound

$$n^E\le C\qquad(n\ge2L). \tag{8}$$

All constants are rational and can be compared exactly. For example, any integer $B\ge2L$ with $B^E>C$ excludes every row $n\ge B$ having this position pattern.

The critical position sums

For $i=3s\in\{27,30,33\}$, set $k=2s-1$. Every $m_t=4s-2-t$ is positive. If $|T|=s$, direct summation gives

$$E=\sum_{t\notin T}m_t-D =\sum_{t\in T}t-\frac{s(s+1)}2. \tag{9}$$

Thus (8) handles every pattern whose position sum exceeds $s(s+1)/2$. The constants in this calculation are:

The equality patterns also have an effective bound. If $\sum T=s(s+1)/2$, then (7) becomes

$$\prod_{p<i}(n/P_p)^{m_{r_p}}\le C. \tag{10}$$

All factors $n/P_p$ are at least one. A set of $s$ distinct nonnegative integers having this sum has maximum at most $2s-1=k$, so $m_{r_p}=2k-r_p\ge k$. Consequently the integer cofactors satisfy

$$\frac{n-r_p}{P_p}\le C^{1/m_{r_p}} \le C^{1/k}<((i-1)!)^2. \tag{11}$$

In particular, write $n-r_2=2^a u$ and $n-r_3=3^b v$. There are only finitely many possibilities for $u,v,r_2,r_3$, all bounded explicitly by

$$1\le u,v<((i-1)!)^2,\qquad 0<|r_3-r_2|\le2s-1,$$

and they satisfy

$$2^a u-3^b v=r_3-r_2. \tag{12}$$

A lower bound for a nonzero linear form in logarithms makes this finite family effective. Equivalently, the effective $S$-part theorem already used in the effectivity note applies to the product of the selected distinct linear factors: all weights in (10) are positive, so any effective positive saving contradicts (10) for sufficiently large $n$. Equation (11) gives a direct route to numerical constants without using that general theorem.

Repeated selected positions are effective as well. In this case the set of selected factors has size at most $s-1$. The degree-reduction argument in the effectivity note gives an elementary bound $U_i(n)\ll_i n^{s-1}$, whereas the existing uniform certificate gives respectively

$$U_{27}(n)\gg n^{243/28},\qquad U_{30}(n)\gg n^{300/31},\qquad U_{33}(n)\gg n^{363/34}.$$

Each exponent exceeds $s-1$. This proves (1) after combining the finitely many effective bounds just described.

Finally write $T=\{t_0<\cdots<t_{s-1}\}$ and $a_q=t_q-q$. The integers $a_q$ are nonnegative and nondecreasing. Condition (1) is equivalent to

$$\sum_{q=0}^{s-1}a_q\le s-1.$$

These patterns are counted by $\sum_{h=0}^{s-1}p(h)$, where $p(h)$ is the partition function, giving 67, 97, and 139. Since $t_{s-1}\le(s-1)+\sum a_q$, the maximum-position bound in (1) follows too. The exact checker verifies these identities and counts; it does not enumerate binomial rows.

What the position restriction leaves open

The remaining list includes $T=\{0,\ldots,s-1\}$. Coordinate-line products cannot eliminate this pattern, even with arbitrary weights. Indeed, write $w_a\ge0$ for the weight of each line $X=a$ and $Y=a$, and let

$$\mu_t=\min_{a+b=t}(w_a+w_b).$$

For $a=0,\ldots,s-1$, use the decompositions

$$s+2a=(s+a)+a,\qquad s+2a+1=a+(s+a+1).$$

They give

$$\sum_{t=s}^{3s-1}\mu_t \le2\sum_{a=0}^{2s}w_a-w_s-w_{2s} \le2\sum_{a\ge0}w_a,$$

which is the degree of the line product. Adding row-diagonal factors adds equally to both sides. Thus this elementary position-sensitive method still needs another arithmetic input on its remaining patterns. In particular, it does not distinguish the assignments of the primes 2 and 3 to the positions, and cannot by itself force their positions to be consecutive.