A342068 a(n) is the smallest k > 1 such that there are more primes in the interval [(k-1)*n + 1, k*n] than there are in the interval [(k-2)*n + 1, (k-1)*n].
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
Offset: 1
Keywords
Examples
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.
Links
- Jon E. Schoenfield, Table of n, a(n) for n = 1..10000
Programs
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Maple
a:= proc(n) uses numtheory; local i, j, k; i:= n; for k do j:= pi(k*n)-pi((k-1)*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 -
Mathematica
a[n_] := Module[{i = n, j, k}, For[k = 1, True, k++, j = PrimePi[k*n] - PrimePi[(k-1)*n]; If[j > i, Break[], i = j]]; k]; Array[a, 100] (* Jean-François Alcover, Jul 01 2021, after Alois P. Heinz *) -
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
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