A224519 - OEIS

%I #11 Jan 19 2019 04:14:59

%S 1,2,3,9,6,7,17,46,45,17,18,19,20,90,106,260,37,77,40,41,42,132,190,

%T 50,51,105,222,58,119,61,62,127,335,70,71,145,74,75,310,326,169,531,

%U 92,93,189,490,101,735,442,113,345,235,854,510,660,271,414,710,438

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

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

%C 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), ...

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

%C 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.

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

%C (1)   p2 - p1 = 72a

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

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

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

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

%p 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:

%Y Cf. A056899, A059100.

%K nonn

%O 4,2

%A _Michel Lagneau_, Apr 09 2013