A339415 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.

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, 150, 2, 14, 22, 26, 34, 46, 54, 90, 126, 162, 174, 10, 14, 34, 46, 50, 62, 54, 90, 126, 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

Offset

1,3

Comments

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.

By symmetry, s and r-s produce the same area; only one of these is included in the table.

The row includes 0 if and only if r/2 is prime, i.e. p is in A339414.

Links

Robert Israel, Table of n, a(n) for n = 1..10020 (rows 1 to 261, flattened)

Formula

The area of the triangle with vertices (p,p), (s,r-s), (q,q) is (q-p)*|p+q-4*s|/4.

Example

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).
The first 10 rows are
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, 150

Maple

R:= 0: count:= 1: q:= 5: nrows:= 1:
printf("0\n"):
while nrows < 20  do
  p:= q; q:= nextprime(q);
  if p+q mod 4 <> 0 then next fi;
  nrows:= nrows+1;
  r:= (p+q)/2;
  T:= select(t -> isprime(t) and isprime(r-t), [$ceil(r/2)..r]);
  count:= count + nops(T);
  V:= map(t -> abs((p-q)*(p+q-4*t)/4), T);
  R:= R, op(V);
  printf("%a\n", V);
od:

Crossrefs

Cf. A098058, A339414.

Sequence in context: A011179 A087570 A124221 * A097625 A371133 A010743

Adjacent sequences: A339412 A339413 A339414 * A339416 A339417 A339418

Keyword

nonn,tabf,look

Author

J. M. Bergot and Robert Israel, Dec 03 2020

Status

approved