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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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2
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7, 11, 23, 109, 211, 307, 1021, 4583, 42967, 297779, 1022443, 1459811, 10781809, 125211211, 11673806759, 3019843939831, 40047392632801, 88212019638251209, 444190204424015227, 57852556614292865039, 9801250757169593701501, 64747502900142088755541, 619216322498658374863033
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OFFSET
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1,1
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COMMENTS
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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.
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LINKS
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FORMULA
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EXAMPLE
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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
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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 *)
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PROG
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(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
(Magma) [n: n in [0..2*10^7] | IsSquare(n^2+NextPrime(n+1)^2-1) and IsPrime(n)]; // Vincenzo Librandi, Aug 02 2015
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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Ulrich Krug (leuchtfeuer37(AT)gmx.de), May 01 2009
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EXTENSIONS
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STATUS
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approved
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