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

Author

Andrzej Kukla, Dec 30 2020

Keywords

• nonn
• more

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.

Examples

			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).
		

References

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

Links

• 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