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%I #11 Dec 04 2020 01:03:13
%S 0,0,2,4,8,0,36,60,4,8,16,0,36,72,84,4,16,20,32,36,72,108,132,54,90,
%T 150,2,14,22,26,34,46,54,90,126,162,174,10,14,34,46,50,62,54,90,126,
%U 198,210,0,144,180,216,240,16,20,40,44,56,64,76,92,14,26,34,50,70,86,94,98,14,98,182,266
%N Table read by rows. If p=A098058(n+1), q is the next prime after p, and r=(p+q)/2, row n consists of the areas (in increasing order) of triangles with vertices (p,p), (s,r-s), (q,q), where s and r-s are prime.
%C If p = A098058(n+1), r is an even number >=4, and Goldbach's conjecture implies that r is the sum of primes s and r-s.
%C By symmetry, s and r-s produce the same area; only one of these is included in the table.
%C The row includes 0 if and only if r/2 is prime, i.e. p is in A339414.
%H Robert Israel, <a href="/A339415/b339415.txt">Table of n, a(n) for n = 1..10020</a> (rows 1 to 261, flattened)
%F The area of the triangle with vertices (p,p), (s,r-s), (q,q) is (q-p)*|p+q-4*s|/4.
%e With p=A098058(5)=17, q=19, r=18, the values of s are 5, 7, 11, 13, corresponding to areas 4, 8, 8, 4 respectively, so row 4 is (4,8).
%e The first 10 rows are
%e 0
%e 0
%e 2
%e 4, 8
%e 0, 36, 60
%e 4, 8, 16
%e 0, 36, 72, 84
%e 4, 16, 20, 32
%e 36, 72, 108, 132
%e 54, 90, 150
%p R:= 0: count:= 1: q:= 5: nrows:= 1:
%p printf("0\n"):
%p while nrows < 20 do
%p p:= q; q:= nextprime(q);
%p if p+q mod 4 <> 0 then next fi;
%p nrows:= nrows+1;
%p r:= (p+q)/2;
%p T:= select(t -> isprime(t) and isprime(r-t), [$ceil(r/2)..r]);
%p count:= count + nops(T);
%p V:= map(t -> abs((p-q)*(p+q-4*t)/4), T);
%p R:= R, op(V);
%p printf("%a\n",V);
%p od:
%Y Cf. A098058, A339414.
%K nonn,tabf,look
%O 1,3
%A _J. M. Bergot_ and _Robert Israel_, Dec 03 2020
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