Factorial Linear Forms and Denominators: Detailed Proofs and Rationality Criteria

Rational grid points and first crossings

Fix $q>0$. To compare a partial sum $H$ with a rational number of denominator $q$, consider the grid $q^{-1}\mathbb{Z}$. Its next point above $H$ is $(\lfloor qH\rfloor +1)/q$. It is at most $S$ exactly when $\lfloor qH\rfloor +1\le qS$. Suppose $H<G\le S$, the number $n!G$ is an integer, and $n!(S-H)<1$. Then

Thus $n!G$ is both the least integer strictly above $n!H$ and the floor of $n!S$. In particular, for $n\ge 2$, suppose a single rational level $G$ satisfies $H_{n+1}<G\le S$ and $n!G\in \mathbb{Z}$. Since $H_n<H_{n+1}$, the tail bounds at both indices give

The next canonical digit is therefore zero. The same level must persist at both indices; choosing an unrelated rational level at each index would not imply this equality.

Set $H_1=0$, the empty partial sum. Now let $\tau \ge 2$ be a first crossing of a rational level $G$, so that $H_{\tau -1}<G\le H_{\tau }$, and write $G-H_{\tau -1}=a/v$ with positive integers $a,v$. The newly added summand gives

The scaled overshoot is

Consequently $-\lfloor \tau !(H_{\tau }-G)\rfloor$ is $0$ or $-1$. Here the minus sign is outside the floor; taking the floor of the negative overshoot would give a different integer at nonintegral overshoots. The displayed integer is $-1$ precisely when the overshoot is at least $1$, or equivalently when

In that case $v\ge \tau !(\tau !-1)$. This extra small-gap condition is not asserted for every crossing. Nor is $-\lfloor \tau !(H_{\tau }-G)\rfloor$ automatically the carry $b_{\tau }$ defined from $Z_{\tau }$: here $G$ is an arbitrary rational level. Neither denominator bound requires $a/v$ to be in lowest terms.

There is also a direct obstruction at prime indices. With ${\mathrm{\Delta }}_m$ as defined in §4, the formal source proves for $m\ge 3$ the criterion

and if $S=a/q$ with $q>0$, then $p\mid Z_p$ for every prime $p>q$, so one exact missed prime $p$ gives $q\ge p$. Exact kernel reduction gives $11\nmid Z_{11}$, hence $q\ge 11$; exact rational normalisation gives $60\nmid Z_{60}$, $64\nmid Z_{64}$ and $67\nmid Z_{67}$, and the prime index $67$ gives the checked bound

The exact-interval census of the short note replaces $67$ by $300000$.

The finite geometric-series identity gives another exact decomposition. For a real $x\ne 0,1$ and an integer $K\ge 0$,

When $x=k!$ and a chosen factorial scale is divisible by $(k!{)}^K$, the scaled finite sum is integral and only the last term retains the factor $k!-1$ in its denominator. The identity isolates one remaining fraction before exact bounding, and it supplies no cofinal family of nonzero residuals.

The earlier record rejects a proposed divisibility strengthening of the first-crossing bound and reports examples at $m=52$ and $m=591$. Its notation for the rational quantity in that claim was not defined, so those reports are not used as a verified counterexample here. The established conclusion remains the lower bound on the denominator’s size, not a specified factor dividing it.

There is a second, more arithmetic mechanism at doubled prime indices, stated in (31). The formal theorem is not restricted to individually computed indices; the divisibility criterion holds for the actual partial sums at every odd prime.