A179479 a(n) is the smallest prime q > a(n-1) such that, for the previous prime p and the following prime r, the fraction (q-p)/(r-q) has denominator 2, for odd n and 1 for even n (or 0, if such a prime does not exist).

3, 5, 7, 11, 13, 17, 19, 29, 43, 53, 67, 71, 79, 97

Offset

1,1

Comments

Conjecture: a(n) > 0 for all n.

Since, as it is accepted in the OEIS, we consider the uncancelled fractions, then, by the condition, for even n, we have (r-q)|(q-p).

Links

Table of n, a(n) for n=1..14.

Example

If n=1, then denominator should be 2. Thus a(1)=3, since (3-2)/(5-3)=1/2. If n=2, then denominator should be 1. Thus a(2)=5, since (5-3)/(7-5)=1/1, etc.

Crossrefs

Cf. A168253, A179210, A179234, A179240, A179256, A179328.

Sequence in context: A212376 A059264 A216484 * A038604 A155026 A295705

Adjacent sequences: A179476 A179477 A179478 * A179480 A179481 A179482

Keyword

nonn

Author

Vladimir Shevelev, Jan 08 2011

Status

approved