

A103271


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


2



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
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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 A320535 A174479
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



