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a(n) = smallest k such that Sum_{i=1..k} log(p_i)/p_i >= n, where p_i is the i-th prime.
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%I #47 Sep 10 2024 08:44:11

%S 3,8,19,43,100,236,562,1354,3300,8119,20136,50302,126451,319628,

%T 811829,2070790,5302162,13621745,35101258,90696900,234924747,

%U 609864582,1586430423,4134442382,10793331294,28221407514,73898377351

%N a(n) = smallest k such that Sum_{i=1..k} log(p_i)/p_i >= n, where p_i is the i-th prime.

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

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

%F a(n) = pi(A347311(n)) = A000720(A347311(n)). - _Michel Marcus_, Sep 06 2021

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

%t Table[k=1;While[Sum[Log@Prime@i/Prime@i,{i,++k}]<n];k,{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)); k; \\ _Michel Marcus_, Sep 06 2021

%Y Cf. A000720, A002387, A016088, A046024, A347311.

%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