Problems & Puzzles: Puzzles

Puzzle 1178  F[n] = Floor[(Prime[n] + 1)^2/(Prime[n + 1] - 1)] Sebastián Martín Ruiz sent the following puzzle: Let: F[n] = Floor[(Prime[n] + 1)^2/(Prime[n + 1] - 1)] I have obtained only one prime on this form:It is 103   n=30  Prime(30)=113 Prime(31)=127 F(30)=103   Q) Find other primes on this form or prove it does not exist   I have tested it from n=1 to 10000000

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From June 22-28, contributions came from Michael Branicky, Alessandro Casini, Jeff Heleen, Emmanuel Vantieghem, Adam Stinchcombe

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Michael wrote:

No further terms for n <= 10^10.

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Alessandro wrote:

Most likely no other primes of this form exist. In fact, letting p[n+1] = p[n] + g, it follows from polynomial division that (p+1)^2 = (p-g+3)(p+g-1) + (g-2)^2. Since the remainder is no-negative, then F[n] = p-g+3 + floor( (g-2)^2/(p+g-1)). If the floor's argument is less than 1, i.e. g^2 - 5g + 5 < p[n], we get that F[n] is even and hence not a prime. The inequality on g is essentially an even weaker condition than Oppermann's conjecture on prime gaps (g^2 < p[n]), for the moment without counterexamples greater than n = 30 (coincidentally) and generally believed to be true. Therefore a new puzzle's prime must be searched for in any hypothetical prime gaps that do not meet the Oppermann's condition, thus certainly n > 4.28 * 10^17.

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Jeff wrote:

I have checked the primes up to 2,000,000,000 with no other solutions found.

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Emmanuel wrote:

Let  f(n)  be  (p(n) + 1)^2 / (p(n+1) - 1).  Thus : F(n) = Floor(f(n)).
 

We can write  p(n)+1  as  p(n+1) - 1 - 2k, where  k  is a natural number <= 0.
 

This allows us to write  f(n)  as :
p(n+1) - 1 - 4k + [4k^2 / (p(n+1) - 1)].
 
For  n < 428472240920394478 and != 30, I found that the last term (the one between [ and ]) is < 1.
For these  n  we thus find  F(n) = p(n+1) - 1 - 4k, even and hence not prime.

I used the table of the 82 known maximal prime gaps from Wikipedia.

If  Oppermann's conjecture is true, the result is true for all  n. 

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Adam wrote:

Write p(n+1)=p and p(n)=p-g, g the "backwards" gap.  Our formula is (p-g+1)^2/(p-1) which expands as [(p-1) - (g-2)]^2/(p-1) = [(p-1)^2 -2(p-1)(g-2) + (g-2)^2] / (p-1) = p-1 - 2(g-2) + (g-2)^2/(p-1) which is an even number + a remainder (g-2)^2/(p-1),  The remainder has to be >=1 so that the floor function adds 1 to make it an odd number.  The belief is that the gap size is related to log(p) and log(p)/p goes to zero fast as p -> 00.  However, (so far) my internet searches have only produced a proved gap (Baker et al) proportional to p^0.525 so a g^2 term could be on the size of p and this remainder could be 1 (maybe even 3 or 5...).
 

Given the test through n=10000000, I think it is unlikely to be prime again.
 

Note: the example when it works has g=14 and p-1=126 so (g-2)^2/(p-1) = 144/126 with a floor of 1.  Then, just luckily, that odd number ends up prime.

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