A218249 - OEIS

A218249

Difference sequence of A219096.

21, 3, 15, 1, 18, 29, 5, 3, 8, 11, 31, 4, 20, 3, 7, 5, 19, 53, 1, 19, 48, 19, 29, 32, 7, 38, 1, 43, 12, 33, 52, 16, 8, 38, 15, 1, 19, 7, 1, 23, 28, 30, 22, 8, 1, 7, 1, 52, 14, 56, 10, 26, 32, 65, 5, 71, 12, 83, 37, 6

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OFFSET

1,1

COMMENTS

Each appearance of 1 represents a prime p for which the next 3 larger primes are p+6, p+12, and p+18. More generally, the sequence gives gap sizes, measured as the number of primes minus 1, between consecutive triples (p, p+6, p+12) and (q, q+6, q+12) of consecutive primes. Conjecture: Every positive integer except 2 occurs infinitely many times.

LINKS

Clark Kimberling, Table of n, a(n) for n = 1..10000

EXAMPLE

A219096 = (15, 36, 39, 54, 55, 73, ...), so that

A218249 = (21, 3, 15, 1, 18, 29, 5, ...).

The first 1 in A218249 represents the primes p(54)=251, p(55)=257, P(56)=263, P(57)=269.

MATHEMATICA

z = 10000; t = Differences[Prime[Range[z]]];

f[n_] := If[t[[n + 1]] - t[[n]] == 0, t[[n]], 0]

u = Table[f[n], {n, 1, 5000}];

p = Flatten[Position[u, 6]] (* A219096 *)

Flatten[Differences[p]] (* A218249 *)