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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
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
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: A357327 A327924 A354057 * A332636 A353742 A238747
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