Before trying the problem "note your opinion as to whether the observed pattern is known to continue, known not to continue, or not known at all."
Let P(n)= The product of the first n primes.
P(n)+1 is like Euclid's method to show there are infinitely many primes, and may or not be prime itself. Now look at the difference between P(n) and the next prime after P(n)+1.
n=1, 5(2)=3
n=2, 11(2x3)=5
n=3, 37(2x3x5)=7
n=4, 223(2x3x5x7)=13
n=5, 2333(2x3x5x7x11)=23
n=6, 30047(2x3x5x7x11x13)=17
n=7, 510529(2x3x5x7x11x13x17)=19
n=8, 9699713(2x3x5x7x11x13x17x19)=23
Are these differences always prime?
Prediction: Yes, almost certainly.
Result: Well, Charlie didn't find a counterexample.
Why it might be true: Consider how the sequence is constructed, for some small P(n) = p1*p2*p3*...p(n). By construction, P(n) is divisible by every prime used to generate the sequence, so with the exception of P(n)+1 none of the next n numbers can be prime; nor can P(n)+2k, P(n)+3k, P(n)+5k... etc.
In fact, by construction, we know that the very next prime is, say, Q. Plainly Q is not divisible by 2,3,5,...p(n) or it wouldn't be prime. Equally, neither is QP(n), since P(n) is divisible by all those small primes. So either QP(n) is prime, or it is the product of factors either of which is larger than n.
We call the gap between p(n)+1 and Q, G. We identify the position of p(n)+1 as 1, then p(n)+2 = 2, p(n)+3 = 3, etc. As in the game Mastermind, if each prime up to n divides the position 1,2,3..., then it is a factor of the corresponding entry, warranted not only to divide the entry but to do so in the correct position:
210 2*3*5*7 Position
211 P 1
212 2*2*53 2
213 3*71 3
214 2*107 4
215 5*43 5
216 2^3*3^3 6
217 7*31 7
But for 210, G=13. Since G is greater than 11, 11 must appear as a factor of at least one entry (in fact 220= 2^2*5*11). We can be sure 11 will appear as a factor of at least one entry, but we don't know which; correct, but not necessarily in the correct place. What number appears in its stead? In position 11 is 221 = 13*17, and 13 is the largest least factor in the gap, and also 13 = G.
So in addition to the warranted positions that are already accounted for, the next few primes between n and G must also be accommodated either as a cofactor in a warranted position, or in one of the open prime positions. E.g for 2310:
2310 2*3*5*7*11
2311 P 1
2312 2*2*17*17 2
2313 3^2*257 3
2314 2*13*89 4
2315 5*463 5
2316 2^2*3*193 6
2317 7*331 7
2318 2*19*61 8
2319 3*773 9
2320 2^4*5*29 10
2321 11*221 11
2322 2*3^3*43 12
2323 23*101 13
2324 2^2*7*83 14
2325 3*5^2*31 15
2326 2*1163 16
2327 13*179 17
2328 2^3*3*97 18
2329 17*137 19
2330 2*5*233 20
2331 3^2*7*37 21
2332 2^2*11*53 22
2332 P 23
The positions up to 12 and the compound ones thereafter are all warranted. Position 13 = 2323 23*101, 17 = 13*179, 19 = 17*137, and position 23 is prime. 23 is the largest least factor in the gap, and also 23 = G.
30030 is a little different because 30031 is not prime. In a way, this is simpler, as every position up to G is warranted:
30030 2*3*5*7*11*13
30031 59*509 1
30032 2^4*1877 2
30033 3^2*47*71 3
30034 2*15017 4
30035 5*6007 5
30036 2^2*3*2503 6
30037 7^2*613 7
30038 2*23*653 8
30039 3*17*19*31 9
30040 2^3*5*751 10
30041 11*2731 11
30042 2*3^2*1669 12
30043 13*2311 13
30044 2^2*7*29*37 14
30045 3*5*2003 15
30046 2*83*181 16
30047 P 17
So yes, it looks like a good bet that such differences are always prime.
Edited on September 22, 2017, 1:58 am

Posted by broll
on 20170921 22:47:03 