Factorial Linear Forms and Denominators: Detailed Proofs and Rationality Criteria

Examples and support refinements

For $D=4$ we need a common divisor of the entire coefficient tail, as well as a finite combination attaining it. Put ${\mathrm{\Lambda }}_n=\mathrm{lcm}(1,\dots ,n)$. We first prove ${\mathrm{\Lambda }}_n\mid u_n$. For $d\mid n$ and a prime $r$, Legendre’s formula gives

Every summand is nonnegative. If $d<r^j\le n$, the second floor is zero and the first is at least one. Counting these powers gives $v_r(W_{d,n})\ge v_r({\mathrm{\Lambda }}_n)-v_r({\mathrm{\Lambda }}_d)$, and therefore ${\mathrm{\Lambda }}_n\mid W_{d,n}{\mathrm{\Lambda }}_d$. Starting from $u_2=2$, induction in the recurrence proves ${\mathrm{\Lambda }}_n\mid u_n$.

Thus $60\mid u_n$ for all $n>4$. Conversely, the finite calculation $u_6=-180$, $u_8=-4200$ gives $23u_6-u_8=60$. The gcd is consequently $G_4=60$. Since $L_4=115$ and $a_4=-55$, we obtain

The vector $12K_4+253U_6-11U_8$ attains this value: its coefficient at index $1$ is $12(-55)+253(-180)-11(-4200)=0$.

The five-term vector used in the short paper is obtained from the same coordinates. The equality $12(-55)-27(-180)-4200=0$ first gives $12K_4-27U_6+U_8$. Its coefficients at indices four and seven are removed by $-26U_5+8U_7$, respectively. Both corrections leave the first coordinate zero. Consequently

The coefficients of the $U_n$ in this expansion sum to $-44$. Since $H_4=143/115$, its remainder is $1380(S-H_4)-44=1380S-1760$. The exact interval calculation appears in §B.6. Thus the coordinates give both the original linear form and the cancellations, independently of a search over supports.

If support is additionally restricted to $n\le 6$, only $u_5=0$ and $u_6=-180$ remain available in (3). The least positive value of $M$ is then $115\cdot 180/gcd(180,55)=4140$. Thus a restriction on the largest supported index can increase the least attainable value of $M$. The following refinements specify the support bound in each case.

There is also a direct congruence proof using only the second and fourth weighted sums. For every $n\ge 2$,

The expression is zero for $n=2,3,4$. For $n\ge 4$, it is divisible by $12$: $12\mid n!$ and $6\mid W_{2,n}$, as follows from $W_{2,2r}=r!\prod _{j=1}^r(2j-1)$ for $r\ge 2$ and the corresponding odd-index recurrence. For $n\ge 5$, it is divisible by $5$ because $5\mid n!$ and the denominators removed to form $W_{2,n}$ and $W_{4,n}$ are coprime to $5$. Modulo $23$, $n!\equiv W_{4,n}$ and $46W_{2,n}\equiv 0$. This proves the congruence. Consequently $V_2=V_4=0$ forces $1380\mid M$, without a restriction on the largest supported index. Conversely, $1482e_2-784e_3-136e_5+83e_6-e_8$ satisfies $M=1380$ and $V_2=V_3=V_4=0$. Thus imposing the additional equation $V_3=0$ does not change the possible values of $M$, although $3e_2-e_3$ shows that it changes the set of admissible vectors.

Among integer vectors supported in $\{2,\dots ,8\}$ with $M=1380$ and $V_2=V_3=V_4=0$, this vector has the smallest support. Every such vector has the following coefficients for some $t,x,y\in \mathbb{Z}$. The coefficient of $K_4$ remains $M/L_4=12$; here $t$ is a new free parameter.

Indeed, the values of the expression in (6) at indices $2,\dots ,8$ are

Therefore the dual identity and $M=1380$ give $11=3{\lambda }_6+21{\lambda }_7+238{\lambda }_8$. Modulo $3$ this yields ${\lambda }_8=-1+3t$, then ${\lambda }_6+7{\lambda }_7=83-238t$. Set $x={\lambda }_5$ and $y={\lambda }_7$; solving $V_2=V_3=V_4=0$ for the coefficients at indices $2,3,4$ gives (7). The coefficient matrix is $\left(\begin{array}{ccc}1& 3& 6\\ 2& 1& 4\\ 2& 6& 1\end{array}\right)$, with determinant $55$, and substitution verifies the converse. Indices $2,3,8$ are nonzero for every integral $t$; neither the pair $({\lambda }_4,{\lambda }_5)$ nor $({\lambda }_6,{\lambda }_7)$ can vanish together. Thus five nonzero entries are necessary at this support bound.

The smallest coefficient ${\ell }^1$ norm for support in $\{2,\dots ,8\}$ is $2416$, attained by $({\lambda }_2,\dots ,{\lambda }_8)=(1482,-784,0,-136,-1,12,-1)$. For $t\ge 1$, the absolute values at indices $2,3,8$ alone sum to $6443t-2267\ge 4176$; for $t\le -1$ they sum to $2267-6443t\ge 8710$. Thus a minimizer has $t=0$. The remaining independent minima are $\underset{x}{min}(|-680-5x|+|x|)=136$ at $x=-136$ and $\underset{y}{min}(|83-7y|+|y|)=13$ at $y=12$. For the first, $680\le |-680-5x|+5|x|$ is at most five times the objective. For the second, the objective is at least $83$ when $y\le 0$, at least $17$ when $0\le y\le 11$, and equals $8y-83\ge 13$ when $y\ge 12$. Adding the fixed contribution $1482+784+1=2267$ gives $2416$. Equality forces $t=0$, $x=-136$ and $y=12$, so this norm minimizer is unique. Among the five-term vectors, $t=0$ is again necessary for the least norm; the pair $({\lambda }_6,{\lambda }_7)$ can then have one nonzero entry only when $y=0$, since $7\nmid 83$. The pair $({\lambda }_4,{\lambda }_5)$ has one nonzero entry only at $x=0$ or $x=-136$, and the latter is smaller. Hence the displayed five-term vector is the unique least-norm vector among those of minimum support, with norm $2486$.

The additional equation $V_3=0$ changes the minimum support size and coefficient norm, while leaving the possible values of $M$ unchanged. With only $V_2=V_4=0$, group the seven coefficients by $a={\lambda }_2+3{\lambda }_3$, $b={\lambda }_4+5{\lambda }_5$, $c={\lambda }_6+7{\lambda }_7$ and $d={\lambda }_8$. Solving $V_2=V_4=0$ and $M=1380$ gives

Each coordinate is nonzero for integral $t$, so four nonzero coefficients are necessary and sufficient. One four-term witness is $-290e_3-136e_5+83e_6-e_8$, with norm $510$. For the coefficient norm on the same set of indices, $|{\lambda }_2|+|{\lambda }_3|\ge |a|/3$ and $|{\lambda }_4|+|{\lambda }_5|\ge |b|/5$; when $t\ne 0$, these bounds together with $|d|$ already sum to at least $794$. At $t=0$, the four groups have unique norm minima $290$, $136$, $13$ and $1$, respectively. Thus the unique norm minimizer under only $V_2=V_4=0$ is $-290e_3-136e_5-e_6+12e_7-e_8$, with norm $440$. For comparison, with $M=1380$ and support contained in $\{2,\dots ,8\}$, the two optimization problems give:

In both rows the norm minimizer has one more nonzero coefficient than a support minimizer. Thus even with $M$ and the allowed indices fixed, least support and least norm need not coincide. These are finite optimization statements, not irrationality conclusions.

There is a uniform restriction $12L_D\mid M$ on support $n\ge 2$. For $n=2,3$ one has $n!=2W_{2,n}$. For $n\ge 4$, both $n!$ and $2W_{2,n}$ are divisible by $12$: if $n=2k$ or $2k+1$, then $W_{2,2k}=k!\prod _{j=1}^k(2j-1)$ is divisible by $6$ for $k\ge 2$. Thus $12\mid M(c)-2V_2(c)$. Every $d!-1$ is coprime to $6$, so $V_2=\cdots =V_D=0$ implies $12L_D\mid M(c)$. This factor is attained at $D=2$ and $D=3$ by $-6e_2+e_4$ and $-6e_2-8e_3+5e_4$, respectively. The next example shows that this need not be the whole restriction.

At $D=6$, the same recurrence gives

To verify the last value, $u_7=0$, while $840=\mathrm{lcm}(1,\dots ,8)$ divides every $u_n$ for $n\ge 8$ by the preceding divisibility. Conversely, $u_8=-4200$, $u_{12}=-14386680$ and $gcd(u_8,u_{12})=840$. Thus $gcd(G_6,a_6)=35$ and the least positive value of $M$ is ${\mu }_6=24L_6$, not $12L_6$. It is attained by $24K_6-16240U_8+U_{12}$: the coefficient at index $1$ is zero and the coefficient at $12$ is $-1$. The finite-gcd formula thus detects the extra factor at depth six, which the uniform divisibility by $12L_D$ misses.