TITLE: Maximal gaps in prime factorizations ("wheel factorization")
QUESTION [6 upvotes]: A wheel factorization is when you remove all the multiples of primes (up to a prime number P) from the product of all primes up to and including P. Examples:
For P=5, you remove all the multiples of 2,3 and 5 from 1 to 2x3x5=30
You are then left with the set {1, 7, 11, 13, 17, 19, 23, 29}
For P=7, you would remove all the multiples of 2,3,5,7 from 1 to 2x3x5x7=210
You are then left with the set {1, 11, 13, 17, ......... 199, 209} etc. 
(Just for the record, I am aware this does NOT generate a set of prime numbers, e.g. 209 = 11x19)
My question is, for a certain P, what is the maximum gap / difference between successive elements of the set and how do I go about proving it? 
When P=3 with the set {1, 5}, the maximum gap is 4
When P=5 with the set {1, 7, 11, 13, 17, 19, 23, 29}, the maximum gap is 6
When P=7 with the set {1, 11, 13, 17, .... 199, 209}, the maximum gap is 10 
When P=11, the maximum gap is 14
I am currently putting together a program to calculate maximum gaps for higher values of P. 
Once I am confident of a pattern, how could I go about proving this (maximum gaps)? I have no idea where to even begin! 
Thanks.

REPLY [2 votes]: Proof that 2 x p_n-1 is not the maximal gap, but the lowest bound of the maximal gap. 
If you take the set of primes {2,3,5,7....p_n-1, p_n)
By the Chinese remainder theorem you can derive an x such that
x = 0 mod N  (where N = 2x3x5x7....p_n-2)
x = 1 mod p_n-1
x = -1 mod p_n
This would create a gap of 2.p_n-1 centred around x. 
Thus the lowest bound of the maximal gap is 2p_n-1.
Comparing 2p_n-1 with the Jacobstahl conjecture it seems they match for the first few pairs as Charles mentioned, and then diverge.