A Fascinating Number Problem
Do there exist 2020 consecutive positive integers satisfying that one is prime and the remaining are composite?
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2 solutions
How are you going to guarantee that you can find n such that n ! + 1 is prime?
Let p be the largest prime smaller than 2 0 2 0 ! + 2 . Then all the numbers from p + 1 to 2 0 2 0 ! + 2 0 2 0 are composite, and there are at least 2 0 1 9 of them. Start with p , and choose the next 2 0 1 9 numbers.
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I was going to make the same comment. The idea in the solution is right - we can find arbitrarily large prime gaps since we know all the numbers from m ! + 2 to m ! + m must be composite for m > 1 . However, the primality of n ! + 1 is tricky; see this OEIS sequence for all n such that n ! + 1 is known to be prime - there are only 4 such n larger than 2 0 1 8 . It seems to still be an open conjecture whether or not there are infinitely many primes of the form n ! + 1 (there are links to papers on this from the OEIS site).
Why not mark hennings?
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As Chris Lewis has said in his comment to me, the OEIS link shows that there are only four values of n > 2 0 1 8 where n ! + 1 is known to be prime. This means that, while your proof works, since such an n exists, it is not at all obvious that it should exist. Just because n ! + 1 is not divisible by 2 , 3 , 4 , . . . , n does not make it prime!
Anyway, as my comment made clear, it is not necessary to know whether n ! + 1 is prime or not.
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I accept your solution is better but my solution is correct as well
Well, as we look for large prime numbers, the difference between consecutive primes also increases, if we just find 2 consecutive large prime numbers whose difference is more than 2018, we are done
Pick a n>2018 such that n!+1 is prime. Consider the sequence: n!, n!+1, n!+2,...,n!+2019. Here, since n>2018 and n!+1 is prime, and we have that all the terms except n!+1 are composite this is the required sequence. Hence the answer is yes.