A103271 a(n) = (prime(n)+prime(n+1)) mod 4.

1, 0, 0, 2, 0, 2, 0, 2, 0, 0, 0, 2, 0, 2, 0, 0, 0, 0, 2, 0, 0, 2, 0, 2, 2, 0, 2, 0, 2, 0, 2, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 2, 2, 2, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 2, 0, 0, 0, 0, 2, 0, 2, 0, 0, 2, 0, 2, 2, 2, 0, 0, 0, 0, 0, 2, 0, 2, 2, 0, 2, 2, 2, 2, 2, 2, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset

1,4

Comments

The number of 2's among the first N terms are: count(10^3) = 381, count (10^4) = 4137, count(10^5) = 42638, count(10^6) = 437423, count(10^7) = 4448503. - M. F. Hasler, Apr 27 2016

In terms of vectors a = (p(n),p(n+1)) mod 4, as considered in the preprint arxiv:1603.03720, the 2's group together the cases a = (1,1) and (3,3) and 0's cumulate cases (1,3) and (3,1). Assuming that the two subcases of each case have roughly the same probabilities, the above counts (i.e., percentage of 44.5% : 55.5% at 10^7) are compatible with the data in the 2nd table on bottom of p.14 where respective percentages vary from 44.8% : 55.1% (at 10^10) to 46% : 54% (at 10^12). I found that at p(n) ~ 10^80, the percentages become closer than 49% : 51%. - M. F. Hasler, May 12 2016

Links

Table of n, a(n) for n=1..106.

R. J. Lemke Oliver and K. Soundararajan, Unexpected biases in the distribution of consecutive primes, arXiv:1603.03720 [math.NT], 2016.

Formula

a(n) = A001043(n) mod 4. - Michel Marcus, Apr 14 2016

Maple

seq(ithprime(n)+ithprime(n+1) mod 4, n=1..150); # Emeric Deutsch, May 31 2005

Mathematica

Table[Mod[Prime@ n + Prime[n + 1], 4], {n, 120}] (* Michael De Vlieger, Apr 27 2016 *)

Prog

(PARI) a(n) = (prime(n) + prime(n+1)) % 4; \\ Michel Marcus, Apr 14 2016

Crossrefs

Cf. A001043, A039702.

Sequence in context: A179212 A105118 A245359 * A029832 A361879 A320535

Adjacent sequences: A103268 A103269 A103270 * A103272 A103273 A103274

Keyword

nonn

Author

Yasutoshi Kohmoto, Jan 27 2005

Extensions

More terms from Emeric Deutsch, May 31 2005

Prepended a(1) = 1, Joerg Arndt, Apr 14 2016

Status

approved