A345328 - OEIS

%I #24 Jun 14 2021 16:04:43

%S 3,20,1096,2981,59874,442413,8886110,65659969,178482301,3584912846,

%T 26489122130,195729609429,3931334297144,78962960182680,

%U 214643579785916,4311231547115195,31855931757113756,86593400423993747,12851600114359308275,34934271057485095348

%N a(n) is the smallest integer k>1 such that |log(k)-round(log(k))| is smaller than 10^(-n).

%C In other words, a(n) is the smallest integer k>1 such that the distance between log(k) and nearest integer to log(k) is smaller than 10^(-n).

%e For n=4 a(n)=2981, because 2981 is the smallest integer greater than 1 such that |log(2981)-round(2981)| = 0.00001409... < 10^(-4).

%p n := 1: for i from 2 to 10^10 do if abs(evalf(log(i)) - floor(log(i) + 1/2)) < 10^(-n) then print(i); n := n + 1 fi end do;

%o (C++) #include <iostream>

%o #include <cmath>

%o using namespace std; int main(int argc, char** argv) {long double n=1; int q=2; for(n=1; n<=9; n++){for(int i=q; i<=100000000000; i++){long double k=log(i); if(abs(k-round(k))<pow(10, -n)){cout<<i<<", "; q=i; break; }}}}

%o (PARI) \\ suitable precision needed.

%o a(n)={my(epsilon=1.0/10^n); for(k=1, oo, my(t=floor(exp(k))); if(k-log(t)<epsilon, while(k-log(t-1)<epsilon, t--); return(t)); if(log(t+1)-k<epsilon, return(t+1)))} \\ _Andrew Howroyd_, Jun 14 2021

%Y Cf. A000193, A000227, A079663, A080021.

%K nonn

%O 1,1

%A _Andrzej Kukla_, Jun 14 2021

%E Terms a(10) and beyond from _Andrew Howroyd_, Jun 14 2021