OFFSET
0,3
COMMENTS
We start with a(0)=0 and a(1)=1, and then the sequence is extended according to these rules:
(1) |a(n+1) - a(n)| = 1 for any n>1,
(2) a(n+1) = a(n-1) iff n is prime.
Is this sequence ultimately positive or ultimately negative or will it change sign indefinitely?
From Ryan Bresler, Jan 04 2021: (Start)
This sequence will contain every integer on "at least one side" of the origin, i.e., it will not have a finite range.
Suppose this sequence has both a finite minimum, R1, and a finite maximum, R2. Since prime gaps become arbitrarily large, we will eventually reach a prime gap g, such that g > R2 - R1. We can see that this prime gap will cause at least one term of this sequence to be outside the interval [R1, R2]. This contradiction shows that all integers on at least one side of the origin will be terms of the sequence.
(End)
LINKS
Rémy Sigrist, Table of n, a(n) for n = 0..10000
FORMULA
a(prime(n)) = prime(1) + Sum_{k=1..n-1} A001223(k)*(-1)^k for any n > 0.
a(n+1) = A065358(n) + 1 for any n >= 0. - Rémy Sigrist, Feb 22 2018
EXAMPLE
a(2) is either a(1) + 1 = 2 or a(1) - 1 = 0.
As 1 is not prime, a(2) = a(1+1) != a(1-1) = 0.
Hence, a(2) = 2.
As 2 is prime, a(3) = a(2+1) = a(2-1) = a(1) = 1.
As 3 is prime, a(4) = a(3+1) = a(3-1) = a(2) = 2.
a(5) is either a(4)+1 = 3 or a(4)-1 = 1.
As 4 is not prime, a(5) = a(4+1) != a(4-1) = 1.
Hence, a(5) = 3.
The first terms can be visualized here (peaks correspond to odd-indexed primes, and valleys to even-indexed primes):
. /\ ...
. / \/
. /\ /
. / \/
. /\ /
. /\/ \/
. /
. 2 5 11 17
. 0 3 7 13 19
PROG
(PARI) y=0; slope=+1; for (x=0, 85, print1 (y ", "); if (isprime(x), slope = -slope); y+=slope)
CROSSREFS
KEYWORD
sign
AUTHOR
Rémy Sigrist, Nov 23 2016
STATUS
approved