Factorial Linear Forms and Denominators: Detailed Proofs and Rationality Criteria

Cofactors give the same primitive vector

A determinant formula may appear to offer a different family. For positive starting index, however, its primitive normalization gives exactly the preceding progression vector, up to sign. We first verify the determinant construction, then identify the two formulas. For integers $n,t\ge 0$, put $s_n=((n+2)!{)}^2$ and $i_j=(t+j)s_n$ for $0\le j\le n+1$. This step size matches the recorded construction; the determinant argument only needs a positive step divisible by every $d=2,\dots ,n+2$. Here the factorial-weight formula is also allowed at index $0$; the case $t=0$ will show why a lower support condition is necessary. Form the integer matrix $A$ with first row $(i_j!{)}_j$ and weighted-sum rows $(W_{d,i_j}{)}_j$ for $2\le d\le n+2$.

To see that $detA\ne 0$, divide column $j$ by $i_j!$. Since $d\mid s_n$, the $d$th weighted-sum row becomes $((d!{)}^{-(t+j)s_n/d}{)}_j$. Removing the nonzero factor $(d!{)}^{-t{s}_n/d}$ from that row leaves a Vandermonde matrix with nodes

These nodes are distinct: $(d!{)}^{1/d}$ is strictly increasing, since $d!<(d+1{)}^d$. The Vandermonde determinant is the product of their pairwise differences [32], and is therefore nonzero. None of the row or column factors removed is zero, so the original determinant is nonzero too.

Let $c$ be the cofactor vector of the first row, and let $N_d$ be the determinant obtained by replacing that row with $(W_{d,i_j}{)}_j$. Cofactor expansion gives $M(c)=detA\ne 0$ and $V_d(c)=N_d$. For $2\le d\le n+2$ the replacement duplicates a row, so $V_d(c)=0$. By (1) each $(V_d(c)-M(c))/(d!-1)$ is an integer and vanishes for $d>\underset{j}{max}{i}_j$, so

A family with $\underset{j}{min}{i}_j=t{s}_n\to \mathrm{\infty }$ and ${\mathcal{R}}_{n,t}\notin \mathbb{Z}$ would prove irrationality. Indeed, for a fixed denominator $q$, every entry of the first row is divisible by $q$ once the least support index is at least $q$, so $q\mid detA$. If $S=a/q$, the integer-difference identity would then force ${\mathcal{R}}_{n,t}\in \mathbb{Z}$. Conversely, if $S$ is irrational, that same identity and $detA\ne 0$ make every ${\mathcal{R}}_{n,t}$ irrational. Taking, for example, $n=0$ and $t\to \mathrm{\infty }$ gives the required family. Thus nonintegrality along such a family is equivalent to the original irrationality question, not a weaker conjecture obtained from the construction.

The condition $t{s}_n\to \mathrm{\infty }$ cannot be replaced by unboundedness of $(n,t)$: at $t=0$, the least supported index stays zero. The coefficient at $i_0$ really is nonzero: its cofactor uses columns $1,\dots ,n+1$, and after nonzero row and column scalings it is a Vandermonde determinant on the weighted-sum nodes. The eventual identity $N_d=detA$ determines the tail, but not the finite intermediate sum, whose terms may have either sign. Thus the determinant proves cancellation, but its nonzero value does not prove nonintegrality of the remainder.

For $t\ge 1$, primitive normalisation identifies this vector exactly. Take $D=n+2$, $\ell =s_n$ and $r=t{s}_n$ in the progression construction of §1.4, and put $N=r+(D-1)\ell$. For any coefficient vector supported on these $D$ indices, the equations $V_2=\cdots =V_D=0$ say that the polynomial

vanishes at the $D-1$ distinct points ${\alpha }_d^{-1}=(d!{)}^{-\ell /d}$, $2\le d\le D$. Its degree is at most $D-1$, so it is a scalar multiple of $\prod _{d=2}^D({\alpha }_{d}X-1)$. Thus the rational solution space is one-dimensional. The progression vector has coefficient $1$ at $N$, so every integer solution is its integer multiple, with multiplier $c_N$. In particular, the cofactor gcd is $|c_N|$, and the primitive cofactor vector is the progression vector up to sign. Choosing the sign so that $M>0$ gives

For example, $n=0,t=1$ gives support $\{4,8\}$, cofactor vector $2520e_4-6e_8$, and primitive vector $-420e_4+e_8$ with $M>0$.

The factorial divisibility proved for the progression coefficient $M$ therefore also applies to this primitive vector. Normalisation still leaves the nonintegrality condition equivalent to irrationality along families whose least support index tends to infinity: $M$ is nonzero, every factorial below that index divides it, and (1) gives the integer-difference identity. This identification does not estimate the distance from an integer. The case $t=0$ is excluded from the identification with the stated progression construction, whose indices are at least $2$.