Concept
Why total-order truncation preserves the integer answer
Before truncation, a product term with indices j_1,...,j_m contains total h-power s=sum j_i. At integer arguments each h contributes a factor of two, so the term is divisible by 2^s. Discarding all s>=ell changes the sum of coefficients only by a multiple of 2^ell. The untruncated sum is the product of the parity indicators. Thus S_ell(product_i B_i mod V^ell) is congruent to one exactly when every L_i is even, and zero otherwise. For ell=2 and two factors, the omitted cross-term contains h(L_1)h(L_2) and is divisible by four. No integrality claim is made for intermediate rational substitutions.
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Updated 2026-11-07
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Math