A224519 For n >= 4, a(n) = (A056899(n) - A056899(n-1))/72, where A056899 lists the primes of the form k^2 + 2.

1, 2, 3, 9, 6, 7, 17, 46, 45, 17, 18, 19, 20, 90, 106, 260, 37, 77, 40, 41, 42, 132, 190, 50, 51, 105, 222, 58, 119, 61, 62, 127, 335, 70, 71, 145, 74, 75, 310, 326, 169, 531, 92, 93, 189, 490, 101, 735, 442, 113, 345, 235, 854, 510, 660, 271, 414, 710, 438

Offset: 4

Author

Michel Lagneau, Apr 09 2013

Keywords

• nonn

Comments

For n >= 3, all prime of the form n^2 + 2 (A056899) are equal to 11 modulo 72.

Observation: this sequence contains couples of consecutive numbers: (2,3), (6,7), (17, 18), (18, 19), (19, 20), (40, 41), (41, 42), ..., (1238, 1239), (1272, 1273), ...

Conjecture: the number of couples such that a(n+1) = a(n) + 1 is infinite.

Consequence: there exists an infinity of triples with 3 successive primes p1 < p2 < p3 of the form n^2 + 2 such that p2 = (p1 + p3)/2 - 36.

Proof: if the conjecture is true, a(n+1) - a(n) = 1 =>

(1) p2 - p1 = 72a

(2) p3 - p2 = 72(a+1)

and (2) - (1) => p2 = (p1 + p3)/2 - 36.

The triples of primes are (11, 83, 227), (83, 227, 443), (1091, 1523, 2027), (11027, 12323, 13691), ...

Examples

			a(5) = (A056899(5) - A056899(4))/72 = (227 - 83)/72 = 2.
		

Crossrefs

Cf. A056899, A059100.

Programs

Maple

with(numtheory): T:=array(1..100):k:=0:for n from 3 to 2000 do: if type(n^2+2,prime)=true then k:=k+1:T[k]:=n^2+1:else fi:od:for i from 1 to k do: printf(`%d, `,(T[i+1]-T[i])/72):od: