A224996 a(n) = floor(1/f(x^(1/n))) for x = 2, where f computes the fractional part.

2, 3, 5, 6, 8, 9, 11, 12, 13, 15, 16, 18, 19, 21, 22, 24, 25, 26, 28, 29, 31, 32, 34, 35, 37, 38, 39, 41, 42, 44, 45, 47, 48, 49, 51, 52, 54, 55, 57, 58, 60, 61, 62, 64, 65, 67, 68, 70, 71, 73, 74, 75, 77, 78, 80, 81, 83, 84, 86, 87, 88, 90, 91, 93, 94, 96, 97

Offset: 2

Author

T. D. Noe, Apr 26 2013

Keywords

• nonn
• easy

Comments

First denominator of continued fraction representing 2^(1/n): [1,a(n),....] so that 1+1/a(n) is first convergent for 2^(1/n). - Carmine Suriano, Apr 29 2014

a(n) is the largest integer y that satisfies (y+1)^n - y^n >= y^n, or equivalently (y+1)^n >= 2*y^n. - Charles Kusniec, Jan 19 2025

Links

• Melvyn B. Nathanson, On the fractional parts of roots of positive real numbers, Amer. Math. Monthly, 120 (2013), 409-429.

Crossrefs

Cf. A224995, A224997, A224998, A001651, A047211, A047203, A047290, A047332.

Cf. A078607 (the smallest integer y that satisfies (y+1)^n - y^n < y^n).

Programs

Mathematica

th = 2; t = Table[Floor[1/FractionalPart[th^(1/n)]], {n, 2, 100}]

Formula

a(n) = floor(n/log(2)-1/2). - Andrey Zabolotskiy, Dec 01 2017