|
| |
|
|
A166332
|
|
Number of primes in (n^(3/2)*(log(n))^(1/2)..(n+1)^(3/2)*(log(n+1))^(1/2)] semi-open intervals, n >= 1.
|
|
3
|
|
|
|
1, 2, 1, 2, 2, 1, 2, 1, 3, 1, 2, 3, 2, 1, 3, 3, 1, 3, 2, 3, 3, 2, 2, 2, 4, 2, 3, 4, 1, 4, 1, 4, 2, 4, 2, 3, 4, 4, 2, 4, 3, 1, 3, 4, 4, 4, 4, 3, 3, 3, 4, 3, 3, 3, 5, 4, 4, 2, 3, 3, 5, 3, 5, 5, 4, 4, 2, 3, 4, 5, 3, 5, 5, 2, 3, 2, 5, 5, 6, 3, 4, 5, 6, 3, 4, 4, 4, 4, 5, 2, 5, 5, 3, 3, 6, 5, 3, 6, 6, 3, 3, 4, 5, 5, 5
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
COMMENTS
|
Number of primes in (n*(n*log(n))^(1/2)..(n+1)*((n+1)*log(n+1))^(1/2)] semi-open intervals, n >= 1.
The semi-open intervals form a partition of the real line for x > 0, thus each prime appears in a unique interval.
a(n) = pi((n+1)^(3/2)*(log(n+1))^(1/2)) - pi(n^(3/2)*(log(n))^(1/2)) since the intervals are semi-open properly.
The n-th interval length is: ~ (1/2)*(n+1/2)^(1/2)*(3*(log(n+1/2))^(1/2)+(log(n+1/2))^(-1/2)) ~ (3/2)*n^(1/2)*(log(n))^(1/2) as n goes to infinity.
The n-th interval prime density is: ~ 2/(3*log(n+1/2)+log(log(n+1/2))) ~ 2/(3*log(n)) as n goes to infinity.
The expected number of primes for n-th interval is: ~ (n+1/2)^(1/2)*(3*(log(n+1/2))^(1/2)+(log(n+1/2))^(-1/2))/ (3*log(n+1/2)+log(log(n+1/2))) ~ n^(1/2)/(log(n))^(1/2) as n goes to infinity.
Using Excel 2003, for n in [1..1123], I obtain a(n) >= 1 (at least one prime per interval).
CAUTION: I will submit the b-file, but since Excel 2003 is limited to 15-digit precision, the rounding might assign to the wrong interval a prime which is extremely close to the limit of 2 successive intervals. The b-file NEEDS TO BE VERIFIED using interval arithmetic! (SEE NEXT)
CAUTION (ADDENDA): for n in [1..1123], the minimum ratio of... ABS(n^(3/2)*(log(n))^(1/2)-ROUND(n^(3/2)*(log(n))^(1/2)))/(n^(3/2)*(log(n))^(1/2)) that I got is 5.04999E-09 which is well above 1E-15 (15-digit limit of Excel 2003), so no interval ended too close to an integral value and every prime has then been assigned to its proper interval. My b-file should then be reliable.
If it can be proved that each interval always contains at least one prime, this would constitute shorter intervals than A143898(n) as n gets large.
The sequence A166363 gives even shorter intervals that seem to always contain at least one prime (for n > 1)!
|
|
|
LINKS
|
|
|
|
CROSSREFS
|
Cf. A144140 (showing that for n^K, K=3/2, some intervals fails to contain primes).
Cf. A166363 (for primes in even shorter intervals).
Cf. A014085 (for primes between successive squares).
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
