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a(n) = floor(1/f(x^(1/n))) for x = 2, where f computes the fractional part.
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%I #30 Jan 20 2025 09:47:15

%S 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,

%T 38,39,41,42,44,45,47,48,49,51,52,54,55,57,58,60,61,62,64,65,67,68,70,

%U 71,73,74,75,77,78,80,81,83,84,86,87,88,90,91,93,94,96,97

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

%C 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

%C 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

%H Melvyn B. Nathanson, <a href="http://dx.doi.org/10.4169/amer.math.monthly.120.05.409">On the fractional parts of roots of positive real numbers</a>, Amer. Math. Monthly, 120 (2013), 409-429.

%F a(n) = floor(n/log(2)-1/2). - _Andrey Zabolotskiy_, Dec 01 2017

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

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

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

%K nonn,easy

%O 2,1

%A _T. D. Noe_, Apr 26 2013