A230666 Primes p such that a Heronian triangle with a fixed side length of 3 contains p as another side length.

5, 149, 29401, 5821349, 8946229758127349, 1771308307978070401, 13748537282247342677718149, 106713369180271088592919391520601, 828287615476676026361062299923143963349, 32470531080787945457870876690417952137154149

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

Table of n, a(n) for n=1..10.

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: A226803 A362494 A332716 * A113560 A094364 A232788

Adjacent sequences: A230663 A230664 A230665 * A230667 A230668 A230669

Keyword

nonn

Author

Frank M Jackson, Oct 27 2013

Status

approved