%I #15 Jun 16 2019 21:13:21

%S 59,739,1601,2777,3041,3307,3911,3917,4787,4987,6199,6317,6959,7669,

%T 8923,10151,12491,14009,14423,14737,15787,16229,16691,17657,24509,

%U 27737,28813,30169,32771,34667,35267,36767,39749,40433,41641,44089,44267,45959,52057,57059,58913,60611,60919,61631,62723

%N Primes prime(k) such that (prime(k), prime(k+1)), (prime(k+2), prime(k+3)), (prime(k+4), prime(k+5)) form a triangle of area 2.

%C If prime(k) is in the sequence, then so is prime(j) if prime(j+2)-prime(j) = prime(k+2)-prime(k), prime(j+4)-prime(j) = prime(k+4)-prime(k), prime(j+3)-prime(j+1)=prime(k+3)-prime(k+1) and prime(j+5)-prime(j+1) = prime(k+5)-prime(k+1).

%C Dickson's conjecture implies that the sequence is infinite.

%H Robert Israel, <a href="/A308649/b308649.txt">Table of n, a(n) for n = 1..10000</a>

%e a(3) = 1601 is in the sequence because 1601, 1607, 1609, 1613, 1619, 1621 are consecutive primes and the triangle with vertices (1601, 1607), (1609, 1613), (1619, 1621) has area 2.

%p P:= map(ithprime, [$2..10000]):

%p g:= (t1,t2,t3,t4,t5) -> t4*(t3-t1)-t2*(t5-t1):

%p P[select(i -> abs(g(seq(P[i+j]-P[i],j=1..5)))=4, [$1..9994])];

%o (PARI) area(u, v, w) = abs((u[1]-w[1])*(v[2]-u[2])-(u[1]-v[1])*(w[2]-u[2]))/2

%o is(p) = my(i=primepi(p), v=primes([p, prime(i+5)])); area([v[1], v[2]], [v[3], v[4]], [v[5], v[6]])==2

%o forprime(p=1, 63000, if(is(p), print1(p, ", "))) \\ _Felix Fröhlich_, Jun 13 2019

%K nonn

%O 1,1

%A _J. M. Bergot_ and _Robert Israel_, Jun 13 2019