A320146 a(n) = 2*prime(n) modulo (prime(n-1) + prime(n+1)).

6, 0, 14, 2, 26, 2, 38, 46, 4, 62, 2, 2, 86, 94, 0, 4, 122, 2, 2, 146, 2, 166, 178, 4, 2, 206, 2, 218, 226, 10, 262, 4, 278, 8, 302, 0, 2, 334, 0, 4, 362, 8, 386, 2, 398, 0, 8, 2, 458, 466, 4, 482, 4, 0, 0, 4, 542, 2, 2, 566, 586, 10, 2, 626, 634, 8, 674, 8, 698, 706, 718, 2, 0, 2, 766, 778, 4, 802, 818, 8, 842

Offset: 2

Author

Andres Cicuttin, Oct 06 2018

Keywords

• nonn

Comments

This sequence has to do with the relative position of primes with respect to their adjacent primes:

(i) if prime(n) is closer to its predecessor than to its successor, then a(n) = 2*prime(n);

(ii) if prime(n) is closer to its successor than to its predecessor, then a(n) = 2*prime(n) - prime(n-1) - prime(n+1); and

(iii) if prime(n) is equidistant from its predecessor and its successor, then a(n) = 0.

Is lim_{n -> infinity} (Sum_{i=1..n} a(i))/(Sum_{i=1..n} prime(i)) finite? If so, what is its value?

Links

• Harvey P. Dale, Table of n, a(n) for n = 2..1000

Crossrefs

Cf. A000040, A001223, A006562, A274263, A276309.

Programs

Maple

seq(modp(2*ithprime(n),(ithprime(n-1)+ithprime(n+1))),n=2..90); # Muniru A Asiru, Oct 07 2018

Mathematica

Table[Mod[2*Prime[n], Prime[n-1] + Prime[n+1]], {n, 2, 120}]
Mod[2#[[2]],#[[1]]+#[[3]]]&/@Partition[Prime[Range[90]],3,1] (* Harvey P. Dale, Jan 03 2019 *)

PARI

a(n) = 2*prime(n) % (prime(n-1) + prime(n+1)); \\ Michel Marcus, Oct 18 2018