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A346151 a(n) is the smallest integer k > 0 such that 1 - tanh(k) < 10^(-n).

%I A346151 #23 Aug 08 2021 01:52:18
%S A346151 1,2,3,4,5,7,8,9,10,11,12,14,15,16,17,18,19,20,22,23,24,25,26,27,28,
%T A346151 30,31,32,33,34,35,37,38,39,40,41,42,43,45,46,47,48,49,50,52,53,54,55,
%U A346151 56,57,58,60,61,62,63,64,65,66,68,69,70,71,72,73,75,76,77
%N A346151 a(n) is the smallest integer k > 0 such that 1 - tanh(k) < 10^(-n).
%C A346151 As k increases, 1 - tanh(k) rapidly approaches 2*exp(-2*k), and the smallest integer k such that 2*exp(-2*k) < 10^(-n), i.e., such that k > (n*log(10) + log(2))/2, is simply ceiling((1/2)*(n*log(10) + log(2))). It seems very likely that this expression gives a(n) for all n >= 0. - _Jon E. Schoenfield_, Jul 08 2021
%e A346151 For n = 3, a(3) = 4 because 4 is the smallest positive integer k such that 1 - tanh(k) < 10^(-3): 1 - tanh(4) = 0.00067....
%t A346151 a[0] = 1; a[n_] := Ceiling @ ArcTanh[1 - 10^(-n)]; Array[a, 100, 0] (* _Amiram Eldar_, Jul 12 2021 *)
%o A346151 (C++)
%o A346151 /* Only suitable for computing a(0) to a(14) due to double precision limits. */
%o A346151 #include <iostream>
%o A346151 #include <cmath>
%o A346151 using namespace std;
%o A346151 int main(int argc, char** argv) {
%o A346151     int control = 1;
%o A346151     for (int n=0; n<=14; n++) {
%o A346151         for (int k=control; k<=100000000; k++){
%o A346151             double x = tanh(k);
%o A346151             double val = abs(1-x);
%o A346151             if (val < pow(10, -n)) {
%o A346151                 cout << k <<",";
%o A346151                 control=k;
%o A346151                 break;
%o A346151             }
%o A346151         }
%o A346151     }
%o A346151 }
%Y A346151 Cf. A346033 (sin), A345670 (cos).
%K A346151 nonn
%O A346151 0,2
%A A346151 _Treanungkur Mal_, Jul 07 2021