Problems & Puzzles: Puzzles

 Puzzle 1181  A puzzle by S. M. Ruiz S. M. Ruiz sent the following puzzle: Let F(n)=Floor(Prime(n+1)^2/Prime(n)) Q1. Prove that if F(n) is prime then: 1) P(n+1)=p(n)+6k , k integer 2) F(n)=Prime(n)+12k There is only one exception n=4 p(4)=7  p(5)=11  F(4)=17 11-7=4    17-7=10 The first examples are: F[n]," ",Prime[n+1]," ",Prime[n]," ",Prime[n+1]-Prime[n]," ",F[n]-Prime[n]   17 11 7 4 10 59 53 47 6 12 73 67 61 6 12 163 157 151 6 12 179 173 167 6 12 223 211 199 12 24 263 257 251 6 12 269 263 257 6 12 283 277 271 6 12 379 373 367 6 12 491 479 467 12 24 Q2. Generalize the result to F(p,q)=Floor(q^2/p) q>p primes

Emmanuel vantieghem wrote on Jiuly 26, 2024:

Q1
Let  f(n) = p(n+1)^2 / p(n)  (so that  F(n) = Floor(f(n)).

Writing  p(n+1)  as  p(n) + g(n)  we have :
f(n) = p(n) + 2g(n) + g(n)^2 / p(n).
If the last term is < 1 then  F(n) = p(n) + 2g(n).
It is a conjecture that this holds for all  n  except when  n = 2, 4, 6, 9, 11, 30.
Of these values only  n = 4  gives prime  F(n)
Now, suppose  n  is such that  g(n)^2 < p(n).
Then, the three numbers
p(n),  p(n+1),  F(n)
form an arithmetic progression with difference  g(n).
If  g(n)  is not divisible by  3  then at least one of the three numbers will be divisible by  3.
So, when  F(n)  is prime, then :
p(n+1) = p(n) + 6v  and  F(n) = p(n) +12v.

Q2
Let  F(p,q)  be  Floor[ q^2 / p]
If  p  is a prime then we may consider the set S(p) of all the primes  q  p,with the property
p <= q < p + Sqrt(p).
We have  S(p) = {p}  for  p = 3, 7, 13, 23, 31, 113  (and conjecturally no other).
The length of the sets  S(p)  varies as follows :

As in part 1  we can prove that for every  q != p  the primality of  F( p,q]  implies
q = p + 6v  and  F(p,q) = p +12v.

For a fixed  p  It might also be interesting to consider the set  T(p) of all primes  q  > p  for which  F(p,q)  is prime.
Then the first computations show :
T(p) = empty  for  p =2, 3, 5, 11
and for all other  p  the set  T(p)  seems to contain many primes.

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