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%I #60 Jan 29 2024 19:10:34
%S 5,13,61,181,421,1741,1861,2521,3121,5101,8581,9661,16381,19801,36721,
%T 60901,71821,83641,100801,106261,135721,161881,163021,199081,205441,
%U 218461,273061,282001,337021,388081,431521,491041,531481,539761,552301
%N Prime hypotenuses of Pythagorean triangles with a prime leg.
%C Apart from the first two terms, every term is congruent to 1 modulo 60 and is of the form 450k^2 +- 30k + 1 or 450k^2 +- 330k + 61 for some k.
%C Every term of the sequence after the second is a prime p congruent to 1 (mod 60), i.e., for n > 2, a(n) is a subsequence of A088955. The Pythagorean triple is {sqrt(2p-1), p-1, p}. - _Lekraj Beedassy_, Mar 12 2002
%C Primes p such that 2*p-1 is the square of a prime. - _Robert Israel_, Sep 16 2014
%C Primes p of the form ((q+1)/2)^2 + ((q-1)/2)^2, where q is a prime; then q belongs to A048161. - _Thomas Ordowski_, May 22 2015
%C The other (i.e., long) leg of the Pythagorean triangle is p-1. - _Zak Seidov_, Oct 30 2015
%H Robert Israel, <a href="/A067756/b067756.txt">Table of n, a(n) for n = 1..10000</a> (first 184 terms from Andreas Boe)
%H H. Dubner and T. Forbes, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL4/DUBNER/pyth.html">Prime Pythagorean triangles</a>, J. Integer Seqs., Vol. 4 (2001), #01.2.3.
%F a(n) = (A048161(n)^2 + 1)/2 = A067755(n) + 1.
%e For a(1)=5, the right triangle is 3, 4, 5 with 3 and 5 prime.
%e For a(10)=5101, the right triangle is 101, 5100, 5101 with 101 and 5101 prime.
%p N:= 10^8: # to get all terms <= N
%p Primes:= select(isprime,[$3..floor(sqrt(2*N-1))]):
%p f:= proc(p) local q; q:= (p^2+1)/2; if isprime(q) then q else NULL fi end proc:
%p map(f, Primes); # _Robert Israel_, Sep 16 2014
%t f[n_]:=((p-1)/2)^2+((p+1)/2)^2; lst={};Do[p=Prime[n];If[PrimeQ[f[p]],AppendTo[lst,f[p]]],{n,6!}];lst (* _Vladimir Joseph Stephan Orlovsky_, Jul 27 2009 *)
%o (PARI) forprime(p=3,10^3,if(isprime(q=(p^2+1)/2),print1(q,", "))) \\ _Derek Orr_, Apr 30 2015
%Y Contains every value of A051859.
%K nonn
%O 1,1
%A _Henry Bottomley_, Jan 31 2002