Erdős 699: full small-prime overlap in the remaining index-six position pattern

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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.

The exact localization lemma permits replacing a selected small-prime power $P_p$ by

$$E_p=p^{\min\{v_p(n-r_p),v_p\binom nj\}}.$$

This can strengthen a polynomial certificate when the column binomial has few carries at one of the selected primes. In the unresolved position pattern

$$(r_2,r_3,r_5)=(2,1,0),$$

the already proved row-zero restriction leaves column residue classes where all three overlaps equal the original prime powers.

Proposition. Suppose $(n,6,j)$ is bad and the selected positions are $(2,1,0)$. Put

$$A_2=v_2(n-2),\qquad A_3=v_3(n-1),\qquad A_5=v_5(n).$$

Then

$$n\equiv250\pmod{300}.$$

For any integer column $j$ with

$$j\equiv11,23,47,\text{ or }59\pmod{60},$$

one has

$$v_p\binom nj\ge A_p\quad(p=2,3,5),\qquad E_2=2^{A_2},\quad E_3=3^{A_3},\quad E_5=5^{A_5}.$$

Here the column must lie in $0\le j\le n$ for the binomial assertion. The overlap conclusion itself only needs the displayed conditions on the row and its three prime-power factors; it does not require badness.

Proof. Maximality at $r_2=2$ forces $A_2\ge2$, hence $n\equiv2\pmod4$. The selected factor at $r_3=1$ is divisible by three, so $n\equiv1\pmod3$. Therefore neither $8\mid n$ nor $9\mid n$. The forced valuation-gap proposition then supplies $25\mid n$, giving $n\equiv250\pmod{300}$.

Each listed column has $j\equiv3\pmod4$, $j\equiv2\pmod3$, and $5\nmid j$. In Kummer's formula

$$v_p\binom nj=\sum_{k\ge1} \mathbf1_{\{j\bmod p^k>n\bmod p^k\}},$$

the first $A_2$ summands at two are all one: at level one the residues are $1>0$, and at every level $2\le k\le A_2$ the row residue is two while the column residue is at least three. At three, every level $1\le k\le A_3$ has row residue one and column residue at least two. At five, every level $1\le k\le A_5$ has row residue zero and nonzero column residue. This proves all three valuation bounds and hence all three equalities for $E_p$. $\square$

These column classes also satisfy the forced row-zero gap in its strongest simple form: $v_5(j)=0$ and $v_5(n)\ge2$. They are closed under column reflection, because $n\equiv10\pmod{60}$ interchanges $11$ with $59$ and $23$ with $47$.

Whether such columns also avoid every common prime at least seven remains unresolved. An argument using the actual overlaps must obtain an additional restriction from those large primes; the selected-position pattern and the known row-zero valuation gap alone provide no power saving in $E_p/P_p$.