OFFSET
1,1
COMMENTS
An infinite number of solutions exists for a^2 + b^2 - 1 = c^2 over the set of natural numbers a, b, c.
If we constrain these to b=a+2, i.e., 2a^2 + 4a + 3 = c^2, the solutions are with a = 1, 11, 69, 407, 2377, ... (The twin prime 11 is also in this sequence here. The solutions can be generated recursively from a(0)=1, m(0)=3 and a(k+1) = 3*a(k) + 2*m(k) + 2, m(k+1) = 4*a(k) + 3*m(k) + 4.)
Filtering these solutions for prime pairs a(k) and b(k) would generate the subset of lower twin primes in the sequence.
The equivalent procedure can be carried out for other prime gaps 2*d such that prime(k)=a, prime(k+1)=a+2*d, 2*a^2 + 4*a*d + 4*d^2 - 1 = m^2. This decomposes the sequence into classes according to the gap 2*d.
a(17) > 5*10^12. - Donovan Johnson, May 17 2010
EXAMPLE
7^2 + 11^2 - 1 = 13^2.
11^2 + 13^2 - 1 = 17^2.
23^2 + 29^2 - 1 = 37^2.
109^2 + 113^2 - 1 = 157^2.
211^2 + 223^2 - 1 = 307^2.
307^2 + 311^2 - 1 = 19^2*23^2.
1021^2 + 1031^2 - 1 = 1451^2.
4583^2 + 4591^2 - 1 = 13^2*499^2.
MATHEMATICA
lst = {}; p = q = 2; While[p < 4000000000, q = NextPrime@ p; If[ IntegerQ[ Sqrt[p^2 + q^2 - 1]], AppendTo[lst, p]; Print@ p]; p = q]; lst (* Robert G. Wilson v, May 31 2009 *)
PROG
(PARI) p=2; forprime(q=3, 1e6, if(issquare(q^2+p^2-1), print1(p", ")); p=q) \\ Charles R Greathouse IV, Nov 06 2014
(PARI) is(n)=issquare(n^2+nextprime(n+1)^2-1)&&isprime(n) \\ Charles R Greathouse IV, Nov 29 2014
(Magma) [n: n in [0..2*10^7] | IsSquare(n^2+NextPrime(n+1)^2-1) and IsPrime(n)]; // Vincenzo Librandi, Aug 02 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
Ulrich Krug (leuchtfeuer37(AT)gmx.de), May 01 2009
EXTENSIONS
Edited and 4 more terms from R. J. Mathar, May 08 2009
a(13) from Robert G. Wilson v, May 31 2009
a(15)-a(16) from Donovan Johnson, May 17 2010
More terms from Jinyuan Wang, Jan 09 2021
STATUS
approved