OFFSET
2,1
COMMENTS
It has been proved that for every integer i > 2 there exists an infinite series of side pairs (j, k) that together with i form a Heronian triangle. It is conjectured that for every prime(n) where n > 1 there exists an infinite series of side pairs (p, q) that together with prime(n) form a Heronian triangle such that either p or q is also prime. See A230666 and A233232 for prime(2) and prime(3). a(n) is the sequence of least such primes for prime(n).
LINKS
Noam Elkies, Heronian triangles with two sides that are prime, Mathoverflow, 2013.
Eugen J. Ionascu, Florian Luca, Pantelimon Stanica, Heron triangles with two fixed sides, arXiv:math/0608185 [math.NT], 2006.
Eugen J. Ionascu, Florian Luca, Pantelimon Stanica, Heron triangles with two fixed sides, Journal of Number Theory, Volume 126, Issue 1, September 2007, Pages 52-67.
Pantelimon Stanica, Santanu Sarkar, Sourav Sen Gupta, Subhamoy Maitra, and Nirupam Kar, Counting Heron triangles with Constraints, Integers, Vol. 13, 2013.
FORMULA
Apart from searching through the first 150000 prime numbers for each prime(n) to form a Heronian triangle (1st Mathematica program), more difficult primes e.g. prime(9)=23 and prime(39)=167 require Pell-type equations to be solved and searched for these least primes (2nd and 3rd Mathematica programs). If a Heronian triangle has side length triples of the form (q, p, p+d) where q = prime(n) and d is odd such that 0 > d > p, then the Pell-type equation is of the form Y^2 - K*X^2 = -J with Y^2 = 4*Area^2/g, X = 2p+d, K = (q^2-d^2)/(4g), J = q^2(q^2-d^2)/(4g) and g = 4 if 16|(q^2-d^2) else g = 1. Other constraints on these primes (see Links) will only permit the following valid pairings:-
prime(n) == 3 mod 4 and a(n) == 1 mod 4
prime(n) == 1 mod 4 and a(n) == 3 mod 4 and prime(n) > a(n)
prime(n) == 1 mod 4 and a(n) == 1 mod 4.
EXAMPLE
a(18)=11 as prime(18)=61, the triple (11, 60, 61) is Heronian and right angled with area=330 and 61 is the least such prime. prime(18)=61==1 mod 4 and a(18)=11==3 mod 4 and prime(18)>a(18).
MATHEMATICA
maxn = 150000; nn=Prime[Range[maxn]]; lst={}; nn1=Prime[Range[2, 100]]; Do[Do[s=(a+b+c)/2; If[IntegerQ[s], area2=s(s-a)(s-b)(s-c); If[area2>0 && IntegerQ[Sqrt[area2]], (AppendTo[lst, b]; Break[])]]; If[b==Prime[maxn], AppendTo[lst, 0]; Break[]], {b, nn}, {a, b-c+2, b+c-2, 2}], {c, nn1}]; lst
(* 1st Program *)
q=23; d=1; nextpair[{y0_, x0_}] := (y=23; x=4; y1=y*y0+x*x0*33; x1=x0*y+y0*x; {y1, x1}); pair=nextpair[{0, q}]; While[!PrimeQ[(pair[[2]]-d)/2] && !PrimeQ[(pair[[2]]-d)/2+d], pair=nextpair[pair]]; primepair={(pair[[2]]-d)/2, (pair[[2]]-d)/2+d}; primepair(* 2nd Program *)
q=167; d=25; y=88751; x=2150; nextpair[{y0_, x0_}] := (If[IntegerQ[(q^2-d^2)/16], k=(q^2-d^2)/16, k=(q^2-d^2)/4]; y1=y*y0+x*x0*k; x1=x0*y+y0*x; {y1, x1}); pair=nextpair[{0, q}]; While[!PrimeQ[(pair[[2]]-d)/2] && !PrimeQ[(pair[[2]]+d)/2], pair=nextpair[pair]]; primepair={(pair[[2]]-d)/2, (pair[[2]]+d)/2}; primepair(* 3rd Program *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Frank M Jackson, Feb 08 2014
STATUS
approved