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 A230666 Primes p such that a Heronian triangle with a fixed side length of 3 contains p as another side length. 3
 5, 149, 29401, 5821349, 8946229758127349, 1771308307978070401, 13748537282247342677718149, 106713369180271088592919391520601, 828287615476676026361062299923143963349, 32470531080787945457870876690417952137154149 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS 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(n-1)-x(n-2)+2, x(0)=1, x(1)=4 starting at x(1). LINKS FORMULA Primes of the form x(m) or x(m)+1 where x(m) is given by x(m) = 6x(m-1)-x(m-2)+2, x(0)=1, x(1)=4 starting at x(1). EXAMPLE 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. MATHEMATICA seq[n_] := seq[n] = Which[n==0, 1, n==1, 4, True, 6seq[n-1]-seq[n-2]+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 = LinearRecurrence[{7, -7, 1}, {1, 4, 25}, 100]; Select[Union[t, t + 1], PrimeQ[#] &] (* T. D. Noe, Dec 09 2013 *) CROSSREFS Cf. A072221. Sequence in context: A075186 A226803 A332716 * A113560 A094364 A232788 Adjacent sequences:  A230663 A230664 A230665 * A230667 A230668 A230669 KEYWORD nonn AUTHOR Frank M Jackson, Oct 27 2013 STATUS approved

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Last modified January 21 12:13 EST 2022. Contains 350477 sequences. (Running on oeis4.)