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A084238 a(2) = 1, and for n > 2 a(n) is the least k such that log(k) < k^(1/n) and log(k-1) >= (k-1)^(1/n). 1

%I #27 Aug 14 2022 23:38:32

%S 1,94,5504,332106,24128092,2099467159,214910065296,25438034785805,

%T 3430631121407802,520643904835474202,87994213187313363255,

%U 16416338625038083857946,3355257076845892674934411,746397968687429806357762425,179698501514006236611711868382,46589028541465014633355926255885

%N a(2) = 1, and for n > 2 a(n) is the least k such that log(k) < k^(1/n) and log(k-1) >= (k-1)^(1/n).

%C A demonstration "that log x increases slower than any power of x. ... No matter how small you make a, the graph of log x is eventually flatter than the graph of x^a."

%D John Derbyshire, Prime Obsession, Bernhard Riemann and the Greatest Unsolved Problem in Mathematics, Joseph Henry Press, Washington, D.C., 2003, Page 72 - 75.

%F For n = 1, a(n) = 1. For n>=2, a(n) = ceiling(e^(-(n+1)*W-1(-1/(n+1)))) where W-1(x) is the Lambert W function with branch -1. - _Joseph C. Y. Wong_, Feb 26 2021

%t Table[ Floor[ FindRoot[ Log[x]^n == x, {x, 10^(2n)}, AccuracyGoal -> 24, WorkingPrecision -> 34][[1, 2]] + 1], {n, 2, 15}]

%K nonn

%O 2,2

%A _Robert G. Wilson v_, May 18 2003

%E a(14)-a(16) from _Joseph C. Y. Wong_, Feb 26 2021

%E Name clarified by _Pontus von Brömssen_, Oct 11 2021

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Last modified April 23 02:23 EDT 2024. Contains 371906 sequences. (Running on oeis4.)