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A293010
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a(n) is the smallest x > 2 to satisfy pi(x-1)/(x-1)^n < pi(x)/x^n, where pi(x) is the prime counting function (A000720).
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0
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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
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1,1
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COMMENTS
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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(x-1)/((x-1)^n) > pi(x)/(x^n), a contradiction. - David A. Corneth, Oct 02 2017
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(x-1) = pi(x)-1 and thus the condition is m-1/(x-1)^n < m/x^n, where m = pi(x). This simplifies to 1-1/m < (1-1/x)^n. A result of Dusart in 1999 shows that x > m(log(m log(m))-1) when m > 1. This implies that (1-1/x)^n > (1-1/(m(log(m log(m))-1)))^n >= 1-n/(m(log(m log(m))-1)) where the last inequality is due to Bernoulli's inequality.
Thus (1-1/x)^n > 1-1/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)
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LINKS
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EXAMPLE
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For n=3, the first integer which satisfies pi(x-1)/((x-1)^3) < pi(x)/(x^3) is 29 = a(3).
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MATHEMATICA
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For[j = 1, j < 11, j++, For[i = 2, i < 1000000 i++, If[(PrimePi[i]/(i^j)) - (PrimePi[i-1]/((i-1)^j)) > 0, Print[i] Break[]]]]
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PROG
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(PARI) a(n) = my(x=3); while(primepi(x-1)/(x-1)^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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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