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A347311
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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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1
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5, 19, 67, 191, 541, 1487, 4079, 11173, 30559, 83137, 226427, 615919, 1675771, 4556771, 12387481, 33677717, 91558231, 248887319, 676566619, 1839125531, 4999337929, 13589640521, 36940536917
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OFFSET
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1,1
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COMMENTS
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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]
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REFERENCES
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Tenenbaum, G. (2015). Introduction to analytic and probabilistic number theory, 3rd ed., American Mathematical Soc. See page 16.
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LINKS
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FORMULA
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EXAMPLE
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a(1) = 5 because log(2)/2 + log(3)/3 + log(5)/5 = 1.034665268989... is the first time the sum is >= 1.
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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