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

Author

Yasutoshi Kohmoto, Jan 27 2005

Keywords

• nonn

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

• Harvey P. Dale, Table of n, a(n) for n = 1..1000
• R. J. Lemke Oliver and K. Soundararajan, Unexpected biases in the distribution of consecutive primes, arXiv:1603.03720 [math.NT], 2016.

Crossrefs

Cf. A001043, A039702.

Programs

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 *)
Mod[Total[#],4]&/@Partition[Prime[Range[120]],2,1] (* Harvey P. Dale, Mar 16 2025 *)

PARI

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

Formula

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

Extensions

More terms from Emeric Deutsch, May 31 2005

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