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%I #10 Mar 21 2017 06:34:11
%S 66,6810,182430,105470250,17356640970,678676246650,1879504308930,
%T 4491035717130,10618004862030,21136679055030,23751520478010,
%U 27081671511090,27596192489190,31721097756750,115248550935750,133303609919430,140838829659930,182797297112430,197799116497230
%N 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.
%C Subsequence of A257049.
%C 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.
%C 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).
%C The corresponding primes p are a subsequence of A056899 (primes of the form n^2+2): 11, 227, 2027, 140627, 4223027, 48650627, 95942027, 171479027, ...
%C We observe that p == 11 mod 72, or p == 11, 83 mod 144. For p>11, p == 27, 227, 627 mod 1000.
%C 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).
%C The following table gives the first values of A, the sides of the triangles and the primes (p-1)/2.
%C +-----------+--------+--------+--------+---------+
%C | A | p | p+2 | 2(p-1)| (p-1)/2 |
%C +-----------+--------+--------+--------+---------+
%C | 66 | 11 | 13 | 20 | 5 |
%C | 6810 | 227 | 229 | 452 | 113 |
%C | 182430 | 2027 | 2029 | 4052 | 1013 |
%C | 105470250 | 140627 | 140629 | 281252 | 70313 |
%C +-----------+--------+--------+--------+---------+
%F a(n) == 6 mod 30.
%e 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.
%p nn:=100000:
%p for n from 1 by 2 to nn do:
%p if isprime(n^2+2) and isprime(n^2+4) and isprime((n^2+1)/2)
%p then
%p printf(`%d, `,n*(2*n^2+4)):
%p else
%p fi:
%p od:
%t 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]
%o (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
%Y Cf. A056899, A085554, A086381, A188158, A253639, A257049.
%K nonn
%O 1,1
%A _Michel Lagneau_, Mar 03 2017
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