OFFSET
1,1
COMMENTS
The sides of each triangle are of the form (k^2+2, k^2+4, 2k^2+2)=(p, p+2,q) where p and p+2 primes (see A257049) => the value of the even side is q = 2k^2+2 and the greatest prime divisor of a(n) is equal to the greatest prime divisor of p-1.
The following table gives the first values (n, a(n)=q, A, p, p+2) where A is the integer area.
+----+--------+--------+------+-------+
| n | a(n)=q | A | p | p+2 |
+-------------+--------+------+-------+
| 1 | 4 | 6 | 3 | 5 |
| 2 | 20 | 66 | 11 | 13 |
| 3 | 452 | 6810 | 227 | 229 |
| 4 | 2180 | 72006 | 1091 | 1093 |
| 5 | 4052 | 182430 | 2027 | 2029 |
| 6 | 6500 | 370614 | 3251 | 3253 |
+----+--------+--------+------+-------+
Note that q = 2*p - 2. - Zak Seidov, May 21 2015
Alternatively, numbers 2k^2 + 2 such that k^2 + 2 and k^2 + 4 are prime. - Charles R Greathouse IV, May 21 2015
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
MATHEMATICA
nn=40000; lst={}; Do[s=(2*Prime[c]-2+Prime[c+1]+Prime[c])/2; If[IntegerQ[s], area2=s (s-2*Prime[c]+2)(s-Prime[c+1])(s-Prime[c]); If[area2>0&&IntegerQ[Sqrt[area2]]&&Prime[c+1]==Prime[c]+2, AppendTo[lst, 2*Prime[c]-2]]], {c, nn}]; Union[lst]
PROG
(PARI) for(k=1, 1e4, if(isprime(k^2+2) && isprime(k^2+4), print1(2*k^2+2", "))) \\ Charles R Greathouse IV, May 21 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Apr 29 2015
STATUS
approved