A326896 Prime numbers p such that 0 < pi(p;10,(9,1)) = pi(p;10,(1,7)) where pi(x;q,(a,b)) is the number of primes p_n <= x such that p_n == a (mod q) and p_(n+1) == b (mod q).

31, 61, 149, 179, 331, 991, 2281, 2351, 2399, 2789, 2999, 3061, 3121, 3181, 5861, 5881, 6089, 6269, 6299, 6359, 6569, 6659, 6971, 7109, 7129, 7451, 7589, 7829, 7949, 8011, 8081, 8219, 8429, 9769, 10039, 10079, 15131, 15881, 15901, 16001, 16229, 16421, 16649, 16729, 16829, 16921, 16979, 17749, 17791

Offset

1,1

Comments

The counting function pi(x;q,(a,b)) is defined by Lemke Oliver and Soundararajan to be the number of primes p_n <= x such that p_n == a (mod q) and p_(n+1) == b (mod q).

Note that it is known that assertions like pi(x;3,1) < pi(x;3,2) or pi(x;4,1) < pi(x;4,3) ("Chebychev's biases") are false infinitely often.

Lemke Oliver and Soundararajan conjectured that pi(x;3,(1,1)) < pi(x,3,(1,2)) and pi(x;4,(1,1)) < pi(x;4,(1,3)) are robust biases, i.e., they always hold.

Here we conjecture that pi(x;10,(9,1)) > pi(x;10,(1,7)) is a semi- robust bias, i.e., it holds from a certain value x_0 on. Here x_0 is conjectured to be 17839. This conjecture seems to be confirmed by the observation that pi(x;10,(9,1)) - pi(x;10,(1,7)) ~ (Li(x)/16) * (1/2) * log(2*Pi/5*log(x)) / log(x).

Note that pi(x;10,(9,1)) = pi(x;5,(4,1)) and pi(x;10,(1,7)) = pi(x;5,(1,2)).

Links

Table of n, a(n) for n=1..49.

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

R. J. Lemke Oliver and K. Soundararajan, Unexpected biases in the distribution of consecutive primes, Proceedings of the National Academy of Sciences of the United States of America, Vol. 113, No. 31 (2016), E4446-E4454.

Example

First prime p such that (p,p') == (9,1) (mod 10) is for p = 29.
First prime q such that (q,q') == (1,7) (mod 10) is for q = 31.
So pi(x;10,(9,1)) = pi(x;10,(1,7)) > 0 first occurs for x = 31, so a(1) = 31.

Prog

(PARI) lista(nn) = {my(vp = primes(nn)); my(nba = 0, nbb = 0); for (n=1, nn-1, my(ok = 0); my(mp = vp[n] % 10); my(mq = vp[n+1] % 10); if ([mp, mq] == [9, 1], nba++; ok=1); if ([mp, mq] == [1, 7], nbb++; ok=1); if (ok && nba && (nba == nbb), print1(vp[n], ", ")); ); } \\ Michel Marcus, Sep 14 2019

Crossrefs

Cf. A030430, A030433, A326897.

Sequence in context: A128470 A195744 A331324 * A132230 A136066 A282326

Adjacent sequences: A326893 A326894 A326895 * A326897 A326898 A326899

Keyword

nonn

Author

A.H.M. Smeets, Sep 13 2019

Status

approved