

A143898


Number of primes between n^K and (n+1)^K, where K = log(1151)/log(95).


8



1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 3, 1, 1, 1, 3, 2, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 3, 2, 1, 2, 3, 2, 1, 3, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 3, 2, 2, 2, 2, 1, 2, 2, 3, 2, 3, 3, 1, 4, 2, 3, 2, 1, 3, 2, 3, 2, 2, 2, 4, 1, 4, 2, 2, 2, 2, 3, 2, 3, 2, 4, 3, 2, 3, 3, 3, 3, 1, 3, 3, 2, 3, 3, 2, 3, 5, 3, 1, 1
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OFFSET

1,2


COMMENTS

This value of K is conjectured to be the least possible such that there is at least one prime in the range n^K to (n+1)^K for n>0. This value of K was found using exact interval arithmetic. For each n <= 110 and for each prime p in the range n to n^1.7, we computed an interval k(n,p) such that p is between n^k(n,p) and (n+1)^k(n,p). The intersection of all these intervals produces a list of intervals. The least value in those intervals is K, which is log(1151)/log(95). We computed 10^5 terms of this sequence to give us confidence that a(n)>0 for all n.
More details about the algorithm: The n^1.7 limit was chosen because we were fairly certain that K would be less than 1.7. Let k(n) be the union of the intervals k(n,p) for p<n^1.7. Then k(n) is the set of exponents e such that the range n^e to (n+1)^e always contains a prime. Let k be the intersection of all the k(n) intervals for n up to N. Then k is the set of exponents e such that there is always a prime in the range n^e to (n+1)^e for n<=N. The number K is the least number in the set k. It appears that as N becomes larger, the set k converges. See A143935. [T. D. Noe, Sep 08 2008]
The constant log(1151)/log(95) is A194362.  John W. Nicholson, Nov 25 2013
1151 counts as a qualifying prime towards both a(94)=1 and a(95)=3, in accordance with use of closed ranges. If prime p were counted only when n^K < p <= (n+1)^K, then term 95 would be 2. If prime p were counted only when n^K <= p < (n+1)^K, then term 94 would be 0. The conjecture in the author's comment implies K is the greatest real value such that for all e <= K there exists n > 0 with no prime p satisfying n^e <= p < (n+1)^e.  Peter Munn, Mar 02 2017
The author's description of the calculation of K implies that K is not an isolated qualifying value; equivalently that K is also the least real value for which there is a positive epsilon such that for all exponent e, K <= e <= K+epsilon and integer n > 0 there is a prime p satisfying n^e <= p <= (n+1)^e. This is a necessary precondition for my Mar 02 2017 deduction from the author's conjecture.  Peter Munn, Aug 21 2019


LINKS

T. D. Noe, Table of n, a(n) for n = 1..10000


MATHEMATICA

k= 1.547777108714197624815033; Table[Length[Select[Range[Ceiling[n^k], Floor[(n+1)^k]], PrimeQ]], {n, 150}] (* T. D. Noe, Sep 08 2008 *)


CROSSREFS

A014085 (number of primes between n^2 and (n+1)^2), both A134034 and A143935 use a larger K.
Sequence in context: A003650 A059233 A327924 * A332636 A238747 A101873
Adjacent sequences: A143895 A143896 A143897 * A143899 A143900 A143901


KEYWORD

nice,nonn


AUTHOR

T. D. Noe, Sep 04 2008, Sep 26 2009, Oct 21 2009


EXTENSIONS

Removed some comments which changed the definition of this sequence.  N. J. A. Sloane, Oct 21 2009


STATUS

approved



