

A293010


a(n) is the smallest x > 2 to satisfy pi(x1)/(x1)^n < pi(x)/x^n, where pi(x) is the prime counting function (A000720).


0



3, 11, 29, 127, 347, 1087, 3079, 8419, 23531, 64553, 175211, 480881, 1304519, 3523901, 9558533, 25874767, 70115473, 189961193, 514272463, 1394193607, 3779851091, 10246935679, 27788566133, 75370121191, 204475052401, 554805820477, 1505578023841, 4086199302077
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OFFSET

1,1


COMMENTS

The integer 2 satisfies the inequality for all values of n (as pi(1) = 0), so it is omitted. With n=0 the sequence is clearly satisfied by all primes.
Conjecture: a(n) exists for all n, that is, for all n, there exists at least one integer which satisfies the inequality.
Occurs when examining convergence of alternating sum to infinity of (1^x)* pi(x)/(x^n).
If a(n) exists it is prime. Proof: If a(n) is composite then pi(x  1) = pi(x), so pi(x1)/((x1)^n) > pi(x)/(x^n), a contradiction.  David A. Corneth, Oct 02 2017
From Chai Wah Wu, Apr 24 2018: (Start)
Conjecture above is true.
Theorem: a(n) exists for all n and satisfies prime(floor(e^W(e^n))) < a(n) < prime(ceiling(e^W(e^(n+1)))), where W is Lambert W function.
Proof: for a fixed n, let x = a(n) if it exists. Since x is prime, pi(x1) = pi(x)1 and thus the condition is m1/(x1)^n < m/x^n, where m = pi(x). This simplifies to 11/m < (11/x)^n. A result of Dusart in 1999 shows that x > m(log(m log(m))1) when m > 1. This implies that (11/x)^n > (11/(m(log(m log(m))1)))^n >= 1n/(m(log(m log(m))1)) where the last inequality is due to Bernoulli's inequality.
Thus (11/x)^n > 11/m if log(m log(m))1 >= n which is satisfied if m >= e^W(e^(n+1)).
The lower bound on a(n) follows analogously from the 1941 upper bound on x due to Rosser: x < m log(m log(m)) when m > 5.
(End)


LINKS

Table of n, a(n) for n=1..28.


EXAMPLE

For n=3, the first integer which satisfies pi(x1)/((x1)^3) < pi(x)/(x^3) is 29 = a(3).


MATHEMATICA

For[j = 1, j < 11, j++, For[i = 2, i < 1000000 i++, If[(PrimePi[i]/(i^j))  (PrimePi[i1]/((i1)^j)) > 0, Print[i] Break[]]]]


PROG

(PARI) a(n) = my(x=3); while(primepi(x1)/(x1)^n >= primepi(x)/x^n, x++); x; \\ Michel Marcus, Oct 02 2017
(PARI) upto(u)=my(t = 1, n = 1, logt = 0, logtm1, logp, logpm1, res = List()); forprime(p = 3, u, t++; logtm1 = logt; logt = log(t); logp = log(p); logpm1 = log(p  1); if(logtm1 + n * logp < logt + n*logpm1, listput(res, p); n++)); res \\ David A. Corneth, Oct 02 2017


CROSSREFS

Cf. A000040, A000720.
Sequence in context: A239713 A122023 A259594 * A236467 A179234 A009183
Adjacent sequences: A293007 A293008 A293009 * A293011 A293012 A293013


KEYWORD

nonn


AUTHOR

Josh Marza, Sep 27 2017


EXTENSIONS

a(20)a(28) from Chai Wah Wu, Apr 24 2018


STATUS

approved



