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%I #38 Dec 16 2023 20:15:08
%S 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,
%T 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,
%U 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
%N a(n) = (prime(n)+prime(n+1)) mod 4.
%C 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
%C 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
%H R. J. Lemke Oliver and K. Soundararajan, <a href="http://arxiv.org/abs/1603.03720">Unexpected biases in the distribution of consecutive primes</a>, arXiv:1603.03720 [math.NT], 2016.
%F a(n) = A001043(n) mod 4. - _Michel Marcus_, Apr 14 2016
%p seq(ithprime(n)+ithprime(n+1) mod 4, n=1..150); # _Emeric Deutsch_, May 31 2005
%t Table[Mod[Prime@ n + Prime[n + 1], 4], {n, 120}] (* _Michael De Vlieger_, Apr 27 2016 *)
%o (PARI) a(n) = (prime(n) + prime(n+1)) % 4; \\ _Michel Marcus_, Apr 14 2016
%Y Cf. A001043, A039702.
%K nonn
%O 1,4
%A _Yasutoshi Kohmoto_, Jan 27 2005
%E More terms from _Emeric Deutsch_, May 31 2005
%E Prepended a(1) = 1, _Joerg Arndt_, Apr 14 2016
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