A160054 Primes prime(k) such that prime(k)^2 + prime(k+1)^2 - 1 is a perfect square.

7, 11, 23, 109, 211, 307, 1021, 4583, 42967, 297779, 1022443, 1459811, 10781809, 125211211, 11673806759, 3019843939831, 40047392632801, 88212019638251209, 444190204424015227, 57852556614292865039, 9801250757169593701501, 64747502900142088755541, 619216322498658374863033

Offset: 1

Author

Ulrich Krug (leuchtfeuer37(AT)gmx.de), May 01 2009

Keywords

• nonn

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

Examples

			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.
		

Crossrefs

Cf. A050791, A129288, A271050.

Programs

Magma

[n: n in [0..2*10^7] | IsSquare(n^2+NextPrime(n+1)^2-1) and IsPrime(n)]; // Vincenzo Librandi, Aug 02 2015

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 *)

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

Formula

{A000040(k): A069484(k)-1 in A000290}.

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