A276812 Prime gap residues mod previous prime gap.

0, 0, 0, 2, 0, 2, 0, 2, 2, 0, 4, 2, 0, 2, 0, 2, 0, 4, 2, 0, 4, 2, 2, 4, 2, 0, 2, 0, 2, 4, 2, 2, 0, 2, 0, 0, 4, 2, 0, 2, 0, 2, 0, 2, 0, 0, 4, 2, 0, 2, 2, 0, 6, 0, 0, 2, 0, 4, 2, 0, 4, 4, 2, 0, 2, 6, 4, 2, 0, 2, 2, 6, 0, 4, 2, 2, 4, 0, 2, 2, 0, 2, 0, 4, 2, 2, 4, 2, 0, 0, 8, 4, 0, 4, 2, 0, 2, 0, 6, 4

Offset: 1

Author

Andres Cicuttin, Sep 18 2016

Keywords

• nonn

Examples

			For n = 4: prime(4+2) = 13, prime(4+1) = 11 and prime(4) = 7. (13-11) % (11-7) = 2 % 4 = 2, so a(4) = 2. - _Felix Fröhlich_, Oct 04 2016
		

Links

• János Pintz, On the ratio of consecutive gaps between primes, arXiv:1406.2658 [math.NT], 2014.
• János Pintz, On the Ratio of Consecutive Gaps Between Primes, in Carl Pomerance and Michael Th. Rassias, Analytic Number Theory; In Honor of Helmut Maier’s 60th Birthday, Springer International Publishing, 2015, ISBN 978-3-319-22239-4, pp. 285-304.

Crossrefs

Cf. A274263, A272863, A274225.

Programs

Mathematica

Table[Mod[Prime[n + 2] - Prime[n + 1], Prime[n + 1] - Prime[n]], {n, 1, 100, 1}]

PARI

a(n) = (prime(n+2)-prime(n+1)) % (prime(n+1)-prime(n)) \\ Felix Fröhlich, Oct 04 2016