Erdős 699: an effective row-zero reduction at every fixed index

P699 research · The effective S-part theorem

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Proposition. Fix $i\ge4$ and $K\ge1$. There is an effectively computable bound for every bad triple with $[n]_{\{p:p<i\}}\le K$.

In particular, above an effective bound depending only on $i$, at least one small prime has a strictly larger valuation in $n$ than in every other numerator factor $n-1,\ldots,n-i+1$.

This is a reduction of possible position patterns, not effective finiteness for an entire remaining index.

Write $S=\{p:p<i\}$ and let $H=i$ when $i$ is prime, $H=1$ otherwise. Let $R_t$ be the part of $n-t$ supported on primes at least $i$. Badness and Lucas's theorem give

$$\frac{R_0}{\gcd(R_0,H)}\mid j.$$

Consequently, if $j/n=a/q$ is reduced, then

$$2\le q\le H[n]_S\le Q:=HK.$$

For every $0\le t<i$, the same localization gives

$$R_t\mid H F_t(a,q),\qquad F_t(a,q)=\prod_{r=0}^t(at-qr).\tag{1}$$

To see this prime by prime, an active prime power dividing $n-t$ forces $j\equiv r$ modulo that power for some $0\le r\le t$. The identity $qj=an$ then gives $at-qr\equiv0$. At the single possible denominator prime $p=i$, an inactive first power is covered by $H$. No inversion of $q$ modulo a prime is needed.

Since $\gcd(a,q)=1$, the polynomial $F_t(a,q)$ is zero exactly when $q\mid t$. For $q=2$ choose $t=1,3$; for $q\ge3$ choose $t=1,2$. Both selected values in (1) are nonzero. As $0<a/q\le1/2$, their product is bounded by $81Q^6$. Thus, for one of the two fixed polynomials

$$f_2(X)=(X-1)(X-2),\qquad f_3(X)=(X-1)(X-3),$$

we have

$$[f_t(n)]_S\ge\frac{f_t(n)}{81H^2Q^6}.\tag{2}$$

BEG's effective Theorem 2.2, already used in the effectivity note, supplies computable $C\ge1$ and $\kappa>0$ such that, for both polynomials,

$$[f_t(n)]_S\le C f_t(n)^{1-\kappa}.$$

Combining this with (2), and using $f_t(n)\ge n^2/4$ for $n\ge6$, gives the effective bound

$$\boxed{n\le2(81CH^2Q^6)^{1/(2\kappa)}.}\tag{3}$$

For the last assertion take $K=\operatorname{lcm}(1,\ldots,i-1)$. If $[n]_S>K$, some $p<i$ has $v_p(n)>\lfloor\log_p(i-1)\rfloor$; otherwise $[n]_S$ divides $K$. For every $1\le t<i$, this forces $v_p(n-t)=v_p(t)<v_p(n)$. So its unique selected maximum is at position zero.

The stronger index-six and index-seven result excludes the complementary position patterns at every admissible row, using exact fixed-slope divisibilities. The present argument supplies the effective reduction uniformly for every fixed $i\ge4$.