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A145913 a(n) = smallest integer k such that 1/log((1 + k)^(1/k)) is bigger than n. 2

%I

%S 1,3,6,10,14,18,22,27,32,37,42,47,52,57,63,68,74,79,85,91,97,102,108,

%T 114,120,126,133,139,145,151,157,164,170,176,183,189,196,202,209,216,

%U 222,229,235,242,249,256,262,269,276,283,290,297,304,310,317,324,331

%N a(n) = smallest integer k such that 1/log((1 + k)^(1/k)) is bigger than n.

%C Note that 1/log((1 + k)^(1/k)) = 1/Hypergeometric2F1(1,1,2,-z).

%C Smallest positive integer k such that k > n*log(1 + k). - _Peter Munn_, Mar 21 2017

%H Charles R Greathouse IV, <a href="/A145913/b145913.txt">Table of n, a(n) for n = 1..10000</a> (first 1000 terms from Indranil Ghosh)

%F a(n) ~ n log n. - _Charles R Greathouse IV_, Mar 24 2017

%t a = {}; k = 1; Do[If[N[((1/n) Log[1 + n])^(-1)] > k, AppendTo[a, n]; k = k + 1], {n, 1, 1010}]; a

%o (PARI) a(n) = {my(k=1); while(1, if((1/log((1 + k)^(1/k))) > n, return (k), k++));};

%o for(n=1, 100, print1(a(n),", ")) \\ _Indranil Ghosh_, Mar 23 2017

%o (PARI) a(n)=my(k=solve(x=n*log(n),n^2,n*log(x+1)-x)\1); while(k <= n*log(k+1), k++); k \\ _Charles R Greathouse IV_, Mar 24 2017

%o (Python)

%o import math

%o def a(n):

%o k=1

%o while True:

%o if (1/math.log((1 + k)**(1/k))) > n: return k

%o else: k+=1

%o print([a(n) for n in range(1, 101)]) # _Indranil Ghosh_, Mar 23 2017

%Y Cf. A145914.

%K nonn

%O 1,2

%A _Artur Jasinski_, Oct 24 2008

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Last modified October 3 16:22 EDT 2022. Contains 357237 sequences. (Running on oeis4.)