A283222 Integer area of integer-sided triangle such that the sides are of the form p, p+2, 2(p-1), where p, p+2 and (p-1)/2 are prime numbers.

66, 6810, 182430, 105470250, 17356640970, 678676246650, 1879504308930, 4491035717130, 10618004862030, 21136679055030, 23751520478010, 27081671511090, 27596192489190, 31721097756750, 115248550935750, 133303609919430, 140838829659930, 182797297112430, 197799116497230

Offset: 1

Author

Michel Lagneau, Mar 03 2017

Keywords

• nonn

Comments

Subsequence of A257049.

The area of a triangle (a,b,c) is given by Heron's formula A = sqrt(s(s-a)(s-b)(s-c)) where its side lengths are a, b, c and semiperimeter s = (a+b+c)/2.

We observe that the sides of each triangle are of the form (k^2+2, k^2+4, 2k^2+2) and Heron's formula gives immediately the area k(2k^2+4) => a(n)= 2*A086381(n)*A253639(n).

The corresponding primes p are a subsequence of A056899 (primes of the form n^2+2): 11, 227, 2027, 140627, 4223027, 48650627, 95942027, 171479027, ...

We observe that p == 11 mod 72, or p == 11, 83 mod 144. For p>11, p == 27, 227, 627 mod 1000.

An interesting property: the greatest prime divisor of a(n) is equal to p. For instance, the prime divisors of 6810 are {2, 3, 5, 227} => p = 227 is the length of the smallest side of the triangle (227, 229, 452).

The following table gives the first values of A, the sides of the triangles and the primes (p-1)/2.

+-----------+--------+--------+--------+---------+

| A | p | p+2 | 2(p-1)| (p-1)/2 |

+-----------+--------+--------+--------+---------+

| 66 | 11 | 13 | 20 | 5 |

| 6810 | 227 | 229 | 452 | 113 |

| 182430 | 2027 | 2029 | 4052 | 1013 |

| 105470250 | 140627 | 140629 | 281252 | 70313 |

+-----------+--------+--------+--------+---------+

Examples

			66 is in the sequence because the area of the triangle (11, 13, 20) is given by Heron's formula with s = 22 and A = sqrt(22(22-11)(22-13)(22-20)) = 66. The numbers 11, 13 and 5 = (11-1)/2 are primes.
		

Crossrefs

Cf. A056899, A085554, A086381, A188158, A253639, A257049.

Programs

Maple

nn:=100000:
for n from 1 by 2 to nn do:
if isprime(n^2+2) and isprime(n^2+4) and isprime((n^2+1)/2)
then
printf(`%d, `,n*(2*n^2+4)):
else
fi:
od:

Mathematica

nn=10000;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 && PrimeQ[(Prime[c]-1)/2], AppendTo[lst,Sqrt[area2]]]], {c,nn}];Union[lst]

PARI

lista(nn) = {forprime(p=2, nn, if (isprime(p+2) && isprime((p-1)/2), ca = p; cb = p+2; cc = 2*(p-1); sp = (ca+cb+cc)/2; a2 = sp*(sp-ca)*(sp-cb)*(sp-cc); if (issquare(a2), print1(sqrtint(a2), ", "));););} \\ Michel Marcus, Mar 04 2017

Formula

a(n) == 6 mod 30.