OFFSET
1,1
COMMENTS
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]
REFERENCES
GĂ©rald Tenenbaum, Introduction to analytic and probabilistic number theory, 3rd ed., American Mathematical Society, 2015. See page 16.
LINKS
Wikipedia, Mertens's Theorems.
FORMULA
a(n) = prime(A347310(n)). - Michel Marcus, Sep 06 2021
EXAMPLE
a(1) = 5 because log(2)/2 + log(3)/3 + log(5)/5 = 1.034665268989... is the first time the sum is >= 1.
MATHEMATICA
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn,more
AUTHOR
N. J. A. Sloane, Sep 06 2021
EXTENSIONS
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
STATUS
approved