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a(n) is the smallest prime q such that Sum_{primes p <= q} log(p)/p >= n.
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%I #38 Sep 10 2024 08:05:02

%S 5,19,67,191,541,1487,4079,11173,30559,83137,226427,615919,1675771,

%T 4556771,12387481,33677717,91558231,248887319,676566619,1839125531,

%U 4999337929,13589640521,36940536917,100415101481,272957090657,741974865617,2016896970001

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

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

%C 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]

%D GĂ©rald Tenenbaum, Introduction to analytic and probabilistic number theory, 3rd ed., American Mathematical Society, 2015. See page 16.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Mertens%27_theorems">Mertens's Theorems</a>.

%F a(n) = prime(A347310(n)). - _Michel Marcus_, Sep 06 2021

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

%t Table[i=1;d=Log@Prime@i/Prime@i;While[d<n,i++;d=d+Log@Prime@i/Prime@i];Prime@i,{n,8}] (* _Giorgos Kalogeropoulos_, Sep 08 2021 *)

%o (PARI) 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

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

%K nonn,more

%O 1,1

%A _N. J. A. Sloane_, Sep 06 2021

%E a(8)-a(16) from _Michel Marcus_, Sep 06 2021

%E a(17)-a(23) from _Jon E. Schoenfield_, Sep 06 2021

%E a(24)-a(27) from _Amiram Eldar_, Sep 10 2024