--- In

primenumbers@yahoogroups.com,

"djbroadhurst" <d.broadhurst@...> wrote:

> Puzzle: (p^2 + 11)/20 is a square for the primes

> p = 3, 13, 67, 4253, 21587, 6950947 ...

> Find the 17th prime in this sequence.

Kevin has now found a reasonably compact answer

as a linear combination of 3 Fibonacci numbers.

In fact the answer can be written in terms of only 2:

Solution [in 19 characters]:

F(2166)/2+3*F(2168)

Proof [by exhaustion, with primality proving]:

P(n,s) = fibonacci(6*n)/2 + 3*fibonacci(6*n+2*s);

{c=0;for(n=0,361,forstep(s=-1,1,2,p=P(n,s);

if(issquare((p^2+11)/20)&&isprime(p),c++;

print1(s*n" "))));print([c]);}

0 -1 1 -3 3 5 -11 15 -45 -47 -65 75 105 253 -303 -359 361 [17]

Comments: Dario Alpern's

http://www.alpertron.com.ar/QUAD.HTM
may be used to prove that my Fibonacci method is exhaustive.

A probable continuation of the sequence of primes is as follows:

F(3966)/2+3*F(3964), 829

F(5334)/2+3*F(5336), 1116

F(8682)/2+3*F(8684), 1816

F(15066)/2+3*F(15068), 3150

F(15438)/2+3*F(15436), 3227

F(41166)/2+3*F(41168), 8604

F(50874)/2+3*F(50876), 10633

F(114702)/2+3*F(114704), 23972

F(117978)/2+3*F(117980), 24657

with the number of digits of each PRP given, after the comma.

The last 3 have been consigned to Henri's repository of PRPs.

David