%I
%S 5,149,29401,5821349,8946229758127349,1771308307978070401,
%T 13748537282247342677718149,106713369180271088592919391520601,
%U 828287615476676026361062299923143963349,32470531080787945457870876690417952137154149
%N Primes p such that a Heronian triangle with a fixed side length of 3 contains p as another side length.
%C The triangle inequality requires that any integer triangle with a fixed side length of 3 can have remaining side lengths of (x, x+1) or (x, x+2). The constraint that primitive Heronian triangles have only one even side will just permit (x, x+1). So a(n) = x or x+1 whenever the Heronian triangle has x or x+1 as a prime and n is the ordered occurrence of this prime. The ordered sequence of x is given by A072221 where x(n) = 6x(n1)x(n2)+2, x(0)=1, x(1)=4 starting at x(1).
%F Primes of the form x(m) or x(m)+1 where x(m) is given by x(m) = 6x(m1)x(m2)+2, x(0)=1, x(1)=4 starting at x(1).
%e a(2)=149 because the triangle with sides (3, 148, 149) is Heronian, 149 is prime and is the 2nd occurrence of such a prime with triangle sides (3, 4, 5) giving 5 as the 1st occurrence.
%t seq[n_] := seq[n] = Which[n==0, 1, n==1, 4, True, 6seq[n1]seq[n2]+2]; lst={}; Do[Which[PrimeQ[seq[m]], AppendTo[lst, seq[m]], PrimeQ[seq[m]+1], AppendTo[lst, seq[m]+1], True, Null], {m, 1, 500}]; lst
%t t = LinearRecurrence[{7, 7, 1}, {1, 4, 25}, 100]; Select[Union[t, t + 1], PrimeQ[#] &] (* _T. D. Noe_, Dec 09 2013 *)
%Y Cf. A072221.
%K nonn
%O 1,1
%A _Frank M Jackson_, Oct 27 2013
