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A340163 For n>=1, smallest integer k such that for all m>=k: m^(1/n)+(m+1)^(1/n) >= (2^n*m+2^(n-1)-1)^(1/n).

%I A340163 #28 Aug 06 2024 06:14:51
%S A340163 0,0,1,2,3,7,14,28,57,115,233,469,945,1902,3823,7680,15420,30948,
%T A340163 62087,124518,249661,500457,1002986,2009771,4026532
%N A340163 For n>=1, smallest integer k such that for all m>=k: m^(1/n)+(m+1)^(1/n) >= (2^n*m+2^(n-1)-1)^(1/n).
%C A340163 For k>1, a(n) <= ceiling(2^(k-3)). This sequence refers to a conjecture, which is a generalization of a Question 723. (iii) from "Collected Papers", Srinivasa Ramanujan.
%D A340163 Srinivasa Ramanujan, Collected Papers, Question 723 in p. 332, Providence RI: AMS / Chelsea (2000).
%H A340163 B. C. Berndt, Y. S. Choi, and S. Y. Kang, <a href="https://citeseerx.ist.psu.edu/pdf/ae75da0be9fb455e2c55daa5fca46ae63e6a60bd">The problems submitted by Ramanujan to the Journal of Indian Math. Soc.</a>, in: Continued fractions, Contemporary Math., 236 (1999), page 14 (see Q723, JIMS VII).
%H A340163 MathOverflow, <a href="https://mathoverflow.net/questions/337758">Generalization of a problem, involving radicals and the floor function, proposed by Ramanujan to the Journal of the Indian Mathematical Society</a>
%e A340163 For n=6, a(6)=7, because for all m<7: m^(1/n)+(m+1)^(1/n) < (2^n*m+2^(n-1)-1)^(1/n) and for all m>=7: m^(1/n)+(m+1)^(1/n) >= (2^n*m+2^(n-1)-1)^(1/n).
%o A340163 (C++)
%o A340163 #include <iostream>
%o A340163 #include <math.h>
%o A340163 using namespace std; int main() {int n=1, k=1; long double a,b; for(n=1; n<18; n++){k=1; while(1) {a=pow(k,1/(long double)n)+pow(k+1,1/(long double)n); b=pow(pow(2,n)*k+pow(2,n-1)-1,1/(long double)n); if(a>=b){cout<<k<<", "; break;} k++;}}}
%K A340163 nonn,more
%O A340163 1,4
%A A340163 _Andrzej Kukla_, Dec 30 2020