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A278603
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A prime mountain: peaks and valleys beyond the origin correspond to prime abscissa (see Comments for precise definition).
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2
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0, 1, 2, 1, 2, 3, 2, 1, 2, 3, 4, 5, 4, 3, 4, 5, 6, 7, 6, 5, 6, 7, 8, 9, 8, 7, 6, 5, 4, 3, 4, 5, 4, 3, 2, 1, 0, -1, 0, 1, 2, 3, 2, 1, 2, 3, 4, 5, 4, 3, 2, 1, 0, -1, 0, 1, 2, 3, 4, 5, 4, 3, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5, 6, 7, 6, 5, 4, 3, 2, 1, 2, 3, 4, 5, 4, 3
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OFFSET
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0,3
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COMMENTS
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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?
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)
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LINKS
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FORMULA
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a(prime(n)) = prime(1) + Sum_{k=1..n-1} A001223(k)*(-1)^k for any n > 0.
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EXAMPLE
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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
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PROG
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(PARI) y=0; slope=+1; for (x=0, 85, print1 (y ", "); if (isprime(x), slope = -slope); y+=slope)
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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