A347310 -id:A347310 - cp's OEIS frontend

A347311 a(n) is the smallest prime q such that Sum_{primes p <= q} log(p)/p >= n.

5, 19, 67, 191, 541, 1487, 4079, 11173, 30559, 83137, 226427, 615919, 1675771, 4556771, 12387481, 33677717, 91558231, 248887319, 676566619, 1839125531, 4999337929, 13589640521, 36940536917, 100415101481, 272957090657, 741974865617, 2016896970001

Offset: 1

N. J. A. Sloane, Sep 06 2021

Suggested by Mertens's theorem that Sum_{p <= x} log(p)/p = log(x) + O(1).

By Mertens's first theorem we have a(n) = exp(n + B3 + o(1)) = e^n * e^B3 * (1 + o(1)) = (3.79081970129... + o(1)) * e^n, where the constant B3 is exp(A083343). Empirically, based on the first 23 terms, it seems plausible that a(n) ~ e^(n + B3) + c*e^(n/2) where c is very roughly -2. - Jon E. Schoenfield, Sep 06 2021 [edited Sep 19 2021, with thanks to Charles R Greathouse IV for his expertise]

			a(1) = 5 because log(2)/2 + log(3)/3 + log(5)/5 = 1.034665268989... is the first time the sum is >= 1.

Gérald Tenenbaum, Introduction to analytic and probabilistic number theory, 3rd ed., American Mathematical Society, 2015. See page 16.

Wikipedia, Mertens's Theorems.

Cf. A002387, A016088, A046024, A083343, A347310.

Table[i=1;d=Log@Prime@i/Prime@i;While[dGiorgos Kalogeropoulos, Sep 08 2021 *)

a(n) = my(k=0, s=0, p=2); while (s < n, s += log(p)/p; k++; p = nextprime(p+1)); prime(k); \\ Michel Marcus, Sep 06 2021

a(n) = prime(A347310(n)). - Michel Marcus, Sep 06 2021

a(8)-a(16) from Michel Marcus, Sep 06 2021
a(17)-a(23) from Jon E. Schoenfield, Sep 06 2021
a(24)-a(27) from Amiram Eldar, Sep 10 2024

cp's OEIS Frontend

A347311 a(n) is the smallest prime q such that Sum_{primes p <= q} log(p)/p >= n.

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