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).

0, 0, 1, 2, 3, 7, 14, 28, 57, 115, 233, 469, 945, 1902, 3823, 7680, 15420, 30948, 62087, 124518, 249661, 500457, 1002986, 2009771, 4026532

Offset

1,4

Comments

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.

References

Srinivasa Ramanujan, Collected Papers, Question 723 in p. 332, Providence RI: AMS / Chelsea (2000).

Links

Table of n, a(n) for n=1..25.

B. C. Berndt, Y. S. Choi, and S. Y. Kang, The problems submitted by Ramanujan to the Journal of Indian Math. Soc., in: Continued fractions, Contemporary Math., 236 (1999), page 14 (see Q723, JIMS VII).

MathOverflow, Generalization of a problem, involving radicals and the floor function, proposed by Ramanujan to the Journal of the Indian Mathematical Society

Example

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).

Prog

(C++)
#include <iostream>
#include <math.h>
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++; }}}

Crossrefs

Sequence in context: A293326 A308092 A262765 * A131666 A135258 A034065

Adjacent sequences: A340160 A340161 A340162 * A340164 A340165 A340166

Keyword

nonn,more

Author

Andrzej Kukla, Dec 30 2020

Status

approved