

A342068


a(n) is the smallest k > 1 such that there are more primes in the interval [(k1)*n + 1, k*n] than there are in the interval [(k2)*n + 1, (k1)*n].


6



2, 6, 10, 5, 3, 7, 5, 3, 5, 5, 7, 4, 11, 3, 5, 5, 7, 3, 4, 3, 7, 4, 5, 5, 5, 6, 6, 9, 3, 6, 8, 4, 6, 5, 7, 5, 5, 6, 5, 5, 7, 4, 9, 6, 4, 10, 3, 3, 4, 4, 7, 4, 6, 4, 5, 5, 4, 5, 4, 8, 6, 7, 7, 5, 10, 6, 3, 3, 6, 4, 4, 4, 4, 4, 4, 9, 8, 6, 6, 6, 3, 5, 6, 5, 6, 5
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OFFSET

1,1


COMMENTS

a(519) is a noteworthy record high value; a(n) < 13 for all n < 519, and a(n) < 19 for all n < 9363 except that a(519)=19.


LINKS

Jon E. Schoenfield, Table of n, a(n) for n = 1..10000


EXAMPLE

The 1st 100 positive integers, 1..100, include 25 primes;
the 2nd 100 positive integers, 101..200, include 21 primes;
the 3rd 100 positive integers, 201..300, include 16 primes;
the 4th 100 positive integers, 301..400, include 16 primes;
the 5th 100 positive integers, 401..500, include 17 primes.
The sequence 25, 21, 16, 16, 17, is nonincreasing until we reach the 5th term, 17, so a(100) = 5.
Considering the positive integers in consecutive intervals of length 519, instead (i.e., [1,519], [2,1038], [3,1557], ...) and counting the primes in each interval, we get a sequence that is nonincreasing until we reach the 19th term, since the 19th interval, [9343,9861], contains more primes than does the 18th, so a(519)=19.


MAPLE

a:= proc(n) uses numtheory; local i, j, k; i:= n;
for k do j:= pi(k*n)pi((k1)*n);
if j>i then break else i:=j fi
od; k
end:
seq(a(n), n=1..100); # Alois P. Heinz, Mar 21 2021


PROG

(Python)
from sympy import primepi
def A342068(n):
k, a, b, c = 2, 0, primepi(n), primepi(2*n)
while a+c <= 2*b:
k += 1
a, b, c = b, c, primepi(k*n)
return k # Chai Wah Wu, Mar 25 2021


CROSSREFS

Cf. A000040, A342069, A342070, A342071, A342839, A342852.
Sequence in context: A320383 A073662 A086553 * A004055 A077933 A144762
Adjacent sequences: A342065 A342066 A342067 * A342069 A342070 A342071


KEYWORD

nonn


AUTHOR

Jon E. Schoenfield, Mar 21 2021


STATUS

approved



