A341952 Let x = (prime(n+1) - prime(n))/2 modulo 3 for n >= 2, then a(n) = -1 if x = 2, otherwise a(n) = x.

1, 1, -1, 1, -1, 1, -1, 0, 1, 0, -1, 1, -1, 0, 0, 1, 0, -1, 1, 0, -1, 0, 1, -1, 1, -1, 1, -1, 1, -1, 0, 1, -1, 1, 0, 0, -1, 0, 0, 1, -1, 1, -1, 1, 0, 0, -1, 1, -1, 0, 1, -1, 0, 0, 0, 1, 0, -1, 1, -1, 1, -1, 1, -1, 1, 0, -1, 1, -1, 0, 1, 0, 0, -1, 0, 1, -1, 1, -1, 1, -1, 1, 0, -1, 0, 1, -1, 1, -1, 0, 1, -1, 1, -1, 0, 0, 1, 0, 0, -1, 0, 0, 1, 0, -1

Offset

2

Comments

Theorem B: apart from the initial term, the subsequence of nonzero terms is alternately +1 and -1 (the subsequence {1, 1} occurs only once, at the start)- for proof see A342156.

Links

Table of n, a(n) for n=2..106.

Formula

a(n) = (A028334(n) mod 3), when 2 -> -1.

a(n) = A102283(A028334(n)).

a(n) = ((prime(n+1) - prime(n))/2 + 1) mod 3 - 1. - Jon E. Schoenfield, Mar 04 2021

Example

a(5)=1 (note that offset is 2, so n-th successive number in this sequence has index n+1) because prime(5+1)=13 and prime(5)=11, then gap=13-11=2, and now gap/2=1 which is congruent to 1 (mod 3).

Mathematica

aa = {}; Do[gap = Prime[n + 1] - Prime[n];
If[Mod[gap/2, 3] == 1, AppendTo[aa, 1],
  If[Mod[gap/2, 3] == 2, AppendTo[aa, -1],
   If[Mod[gap/2, 3] == 0, AppendTo[aa, 0]]]], {n, 2, 106}]; aa
a[n_] := Mod[(Prime[n + 1] - Prime[n])/2, 3, -1]; Array[a, 100, 2] (* Amiram Eldar, Feb 25 2021 *)

Prog

(PARI) a(n) = my(x=(prime(n+1) - prime(n))/2 % 3); if (x==2, -1, x); \\ Michel Marcus, Feb 26 2021

Crossrefs

Cf. A028334, A341765, A342156.

Sequence in context: A350070 A359834 A179775 * A167686 A190207 A156706

Adjacent sequences: A341949 A341950 A341951 * A341953 A341954 A341955

Keyword

sign

Author

Artur Jasinski, Feb 24 2021

Status

approved