OFFSET
2,1
COMMENTS
Conjecture: the sequence contains all primes such that the first appearances of them are in the natural order.
Provided all primes appear, they must appear in the natural order. Suppose otherwise: For A=A256189 and some n and primes p < q, q divides a = A(n) but not A(k) for k < n and p does not divide A(k) for k <= n. But then A(n) should be pa/q. - Danny Rorabaugh, Mar 31 2015
LINKS
Peter J. C. Moses, Table of n, a(n) for n = 2..1001
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, Mar 26 2015
STATUS
approved