|
|
A237518
|
|
Least primes that together with prime(n) forms a Heronian triangle, starting at n = 2.
|
|
1
|
|
|
5, 3, 4729, 13, 5, 17, 37, 5280071830550089, 5, 97, 13, 17, 61, 1824001, 53, 109, 11, 3301, 1009, 19, 241, 241, 17, 11, 29, 409, 6841, 11, 17, 3169, 181, 41, 157, 3, 457, 13, 10369, 231781748893580717709514473745694370721, 173, 277, 19, 7297, 31, 53, 3049, 373
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
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
|
|
|
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
|
|
|
STATUS
|
approved
|
|
|
|