Geometric Moments and Rational Lambert Values: Proofs and Further Results

The range of subset sums.

The residue condition and the real-error condition can be counted together. Within each residue class, subdivide the range of subset remainders into intervals of width $D/n$; a bound on repeated remainder values then forces a nonzero small difference. Fix a positive integer $n$. For $M$ integer rows $(A_j,B_j)$ and a modulus $D\ge 1$, set $e_j=A_{j}F(3/2)-B_j$. For each attained residue vector $b$, let $T_b$ be the difference between the largest and smallest remainders $\sum _j{\epsilon }_j{e}_j$ with ${\epsilon }_j\in \{0,1\}$ and $\sum _j{\epsilon }_j(A_j,B_j)\equiv b(\mathrm{mod}D)$. Let the integer $k_b$ bound the number of subsets attaining any one remainder value in that residue class. Then

produces two subsets with equal residues and distinct remainders less than $D/n$ apart. Within each residue fibre, subtract the least remainder, multiply by $n/D$, and take floors. These integer indices range from $0$ to $\lfloor n{T}_b/D\rfloor$, including a separate final index when $T_b$ is a positive integral multiple of $D/n$. Equal indices give a remainder difference strictly less than $D/n$. If each index corresponded to only one remainder value, it would account for at most $k_b$ subsets, contradicting the displayed inequality. Subtracting the two subset sums and dividing their coefficients by $D$ gives an integer form with nonzero absolute value less than $1/n$. The common width bound $T=\sum _j|e_j|$ and a common multiplicity bound $k$ give the coarser count $Qk(\lfloor nT/D\rfloor +1)$, where $Q$ is the number of attained residue vectors. All rows must have the same normalisation. Primitive rows can still have repeated subset sums, and positivity need not survive subtraction.