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A166968
Minimum k such that for all m >= k there is a prime p with m < p < m * (n+1)/n.
2
2, 8, 9, 24, 25, 32, 33, 48, 115, 116, 117, 118, 118, 140, 140, 141, 200, 212, 212, 213, 294, 294, 318, 318, 319, 319, 320, 320, 320, 524, 525, 525, 526, 526, 526, 527, 527, 528, 528, 1328, 1329, 1330, 1331, 1331, 1332, 1333, 1333, 1334, 1334, 1335, 1335
OFFSET
1,1
COMMENTS
The first term was proved by Chebyshev in 1850: for all m > 1, there is a prime number between m and 2m. It is known by Bertrand's Postulate after Joseph Bertrand, who first conjectured it in 1845, and also by Chebyshev's Theorem.
The result a(5)=25 was proved by Jitsuro Nagura in 1952.
The result a(16597)=2010760 was proved by Pierre Dusart in 1998.
LINKS
Jitsuro Nagura, On the interval containing at least one prime number, Proc. Japan Acad. 28: 177-181, 1952.
Eric Weisstein's World of Mathematics, Bertrand's Postulate.
EXAMPLE
For n=4, there are no primes between 23 and 23*5/4 = 28.75. But, for all m >= 24, there is a prime p such that m < p < 5m/4, so a(4) = 24.
PROG
(PARI) /* This function searches until it finds 10 primes between x and x*(n+1)/n */
pi_excl(y) = if(y==floor(y), primepi(y)-isprime(y), primepi(y)) /* all primes < y, primepi(y) is all primes <= y */
pbetween(x, y) = pi_excl(y) - primepi(x)
A166968(n) = {local(pr, x, r); pr=0; x=1; r=0; while(pr<10, pr=pbetween(x, x*(n+1)/n); if(pr==0, r=x+1); x=x+1); r}
(Sage)
def a_list() :
known_n, known_k = (16597, 2010760)
L = [0] * known_n
L[known_n-1] = known_k
for n in range(known_n-1, 0, -1) :
L[n-1] = 1 + next(k for k in range(L[n]-1, 0, -1) if next_prime(k) >= k*(n+1)/n)
return L
# Eric M. Schmidt, Oct 21 2017
CROSSREFS
Sequence in context: A046681 A259672 A163619 * A075644 A088825 A337706
KEYWORD
nonn
AUTHOR
Michael B. Porter, Oct 25 2009
EXTENSIONS
Edited by Eric M. Schmidt, Oct 21 2017
STATUS
approved