Problems & Puzzles: Puzzles

 Puzzle 630. Gap between p(n-1) & p(n+1) Emmanuel Vantieghem poses the following question: One of the first striking results about primes was the fact that the difference between two consecutive primes could become arbitrary big.  The proof was also surprisingly simple: none of the numbers  n!+2, n!+3, ..., n!+n-1  are primes. But, what about the difference between three consecutive primes?  Is it possible that the differences  p(n)-p(n-1)  and  p(n+1)-p(n) both are arbitrary big ? Q. Prove your answer.

Contributions came from Fred Schneider, Luke Pebody, J. K. Andersen & Emmanuel Vantieghem

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Fred wrote:

"The proof was also surprisingly simple: none of the numbers  n!+2, n!+3, ..., n!+n-1  are primes."
Actually, n! +n must be composite, too.

Similarly, none of n! -2,  n! -3 ...  n! -n are primes (for n>3)

It's clear that for any m*n! where m is a positive integer:  m*n! +/-2, m*n! +/-3 ... m*n! +/- n must all be composite (for n > 3)

If there exists a prime: m*n! -1 OR m*n!+1,  then both the gaps between it and both the previous prime and the next prime can be arbitrarily large.  This is not a proof but there are an infinite number of m's to choose from.  Surely, at least one m for a given n must result in either m*n! -1 OR m*n!+1 being a prime.

This might be an interesting sequence to calculate of minimum m's for each n.

...

I tried out my idea and for each primorial(p)  where p is a prime < 1100, I found that the minimal m where m*primorial(p) +1 or -1 is prime always has m much lower than p.  This is true when taken individually, too. In fact, I want to conjecture that m will always be less than p.

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Luke wrote:

For all integers K, there exists P>K+2 with P prime. Let Q be the product of all primes less than 2P other than P.

Q is coprime with P, so by Dirichlet's Theorem on Primes in an Arithmetic Progression, there exists t>=2 such that Qt+P is prime. However, for all non-zero k with |k|<=K, P+k has a prime factor in common with Q, so Qt+P+k is not prime.

This shows that Qt+P is a prime with a gap of K on both sides.

Proving that there are three consecutive arbitrarily long regions would probably be much harder.

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J. K. Andersen wrote:

The gaps can both be arbitrarily big.
This can for example be proven by showing that for any n,
there is k such k*n#+1 is the only prime from k*n#-n to k*n#+n.
We need a k where k*n#+1 is prime and k*n#-1 is composite.
Let p be the smallest prime above n.
Let b be the smallest natural number such that p divides b*n#-1.
n# and p are coprime so b always exists.
Now consider the arithmetic progression
(a*p+b)*n#+1 = a*(p*n#) + (b*n#+1), where a is the only non-constant.
p*n# and b*n#+1 are coprime so by Dirichlet's theorem,
there are infinitely many primes of form (a*p+b)*n#+1.
The next prime must be above (a*p+b)*n#+n.
p always divides (a*p+b)*n#-1 so the previous prime is below (a*p+b)*n#-n.

Example: If n=11 then p=13 and b=3. 13 always divides (a*13+3)*11#-1.
(a*13+3)*11#+1 is prime for a = 3, 4, 5, 6, 11, 14, 15, ...

A natural question is finding the smallest prime to give at least a given gap.
http://oeis.org/A023186 is:
"Lonely (or isolated) primes: increasing distance to nearest prime."
A023186 shows the terms below 10^11.

http://www.fishmech.net/zipdisks/LONELY.TOT/LONEMAX.SUM continues
to 10^12 and shows the associated gaps.
My computation agrees and adds 4 terms from 10^12 to 10^13:

gap      prime
---  -------------
300  1362810282439
324  1480975873513
328  5551890283531
340  9156364643509

All 4 are also in the b-file for the closely related http://oeis.org/A051650:
http://oeis.org/A051650/b051650.txt

A051650 is: "Lonely numbers: distance to closest prime sets a new record."
Both primes and composites are allowed in that sequence.
Some primes in our problem are not in A051650 because they are beaten
by a prime gap around a smaller composite.

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Emmanuel wrote:

The answer is yes.  I.e. : for any N we can find three consecutive primes  p, q, r  such that q - p > N  and  r - q > N.  The proof runs as follows :

Let  p(n)  denote the n-th prime.  Choose  n  such that  p(n) > N.  Let  P(n)  be the primorial  p(n)# = p(1)*p(2)* ... *p(n).  Choose  m such that  m*P(n) - 1  is divisible by  p(n+1).

Now, consider the arithmetic progression

{ (m+k*p(n+1))*P(n) + 1  |  k = 0, 1, 2, ...  }.

The first term is  m*P(n)+1  and the common difference is  P(n+1).  These two numbers are obviously relatively prime.  Hence, by Dirichlet's theorem on primes in arithmetic progressions, there is a  K  such that  P = (m+K*p(n+1))*P(n) + 1  is prime.  It is easy to see that all the numbers  P+1, P+2, ... , P+p(n)  and  P-1, P-2, ... , P-p(n)  are composite (especially, note that P-2  is divisible by p(n+1) by the choice of  m).  Thus, the prime preceding  P  must be smaller than  P-N  and the prime succeeding  P  must be bigger than  P+N, proving aour assertion.

An immediate consequence is that, whatever  n (>0)  may be, there exists a smallest  w  such that precisely one of the two numbers w*P(n) - 1  and  w*P(n)+1  is prime.  However, this is not the most efficient way to find small  p, q, r  with  q - p > N  and  r - q > N. For instance, when  N = 100, then  n = 26  and the smallest  w  is  2, which leads to the 'middlest' prime

2*P(26)+1 = 465724728716994721800126633761014726141.

However, the smallest triplet  (p, q, r)  with  q-p  and  r-q > 100  is :(72546143, 72546283, 72546391), which is far much smaller ...

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