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A160054
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Primes prime(k) such that prime(k)^2+prime(k+1)^2-1 is a perfect square.
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0
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7, 11, 23, 109, 211, 307, 1021, 4583, 42967, 297779, 1022443, 1459811, 10781809, 125211211, 11673806759, 3019843939831
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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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, ie, 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. [From Donovan Johnson (donovan.johnson(AT)yahoo.com), May 17 2010]
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FORMULA
| {A000040(k): A069484(k)-1 in A000290}.
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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.
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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 [From Robert G. Wilson v (rgwv(AT)rgwv), May 31 2009]
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CROSSREFS
| Cf. A129288, A050791.
Sequence in context: A118072 A181841 A076855 * A027830 A134043 A102373
Adjacent sequences: A160051 A160052 A160053 * A160055 A160056 A160057
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KEYWORD
| nonn
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AUTHOR
| Ulrich Krug (leuchtfeuer37(AT)gmx.de), May 01 2009
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EXTENSIONS
| Edited and 4 more terms from R. J. Mathar (mathar(AT)strw.leidenuniv.nl), May 08 2009
a(13) from Robert G. Wilson v (rgwv(AT)rgwv), May 31 2009
a(15)-a(16) from Donovan Johnson (donovan.johnson(AT)yahoo.com), May 17 2010
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