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A278425
Largest k such that there are no primes between kn and k(n+1); -1 if no such k exists.
1
1, 1, 1, 2, 1, 4, 2, 4, 1, 2, 3, 4, 9, 1, 6, 3, 7, 5, 6, 10, 4, 2, 5, 5, 8, 7, 2, 5, 11, 4, 3, 10, 9, 6, 15, 6, 8, 4, 3, 8, 5, 7, 5, 12, 2, 7, 3, 11, 6, 6, 10, 9, 10, 6, 2, 3, 5, 23, 9, 6, 4, 10, 4, 8, 6, 8, 20, 5, 9, 19, 4, 12, 7, 18, 7, 7, 2, 6, 17, 6, 14, 6, 16, 16, 6, 9, 13, 19, 15, 14, 18, 4, 18, 5, 14, 14, 13, 4, 9, 8
OFFSET
1,4
COMMENTS
This sequence deals with the question of whether there is always a prime between nk and n(k+1). For n<=3 the answer has been proven to be yes (see links and examples). For n>3 the problem remains open, however we can conjecture the values of a(n) by checking the first few hundred k.
Conjecture: For every n, there exists a finite m such that for every k>m there is at least one prime between kn and k(n+1). In other words, a(n) is never -1.
Conjecture follows from the Prime Number Theorem: for fixed n, the number of primes between kn and k(n+1) is asymptotic to k/log(k) as k -> infinity, and in particular is nonzero for all sufficiently large k. - Robert Israel, Nov 28 2016
LINKS
M. El Bachraoui, Primes in the interval [2n,3n], International Journal of Contemporary Mathematical Sciences, volume 1, number 13, pages 617-621, 2006.
Andy Loo, On the primes in the interval [3n,4n], International Journal of Contemporary Mathematical Sciences, volume 6, number 38, pages 1871-1882, 2011.
EXAMPLE
Bertrand's postulate shows that for k>1 there is always a prime between k and 2k. Hence a(1) = 1.
In 2006, M. El Bachraoui showed that for k>1 there is always a prime between 2k and 3k. Hence a(2) = 1.
In 2011, Andy Loo showed that for k>1 there is always a prime between 3k and 4k. Hence a(3) = 1.
CROSSREFS
Cf. A060715.
Sequence in context: A352791 A274455 A153279 * A309019 A082908 A086449
KEYWORD
nonn
AUTHOR
Dmitry Kamenetsky, Nov 28 2016
STATUS
approved