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For n >= 2, a(n) is the least positive integer x such that 2*x^n > (x+2)^n. For example, a(2)=5 as 4^2=16, 5^2=25, 6^2=36 and 7^2=49.

Coincides with floor( 2*n/(log 2) ) for all n from 1 to 777451915729367 but differs at 777451915729368. See A129935.

The first few values of n for which this sequence differs from floor( 2*n/(log 2) ) were found by Dean Hickerson in 2002. - N. J. A. Sloane, Apr 30 2014

The sequence floor( log(n)/(2*log(2)) ) is mentioned by Erdős and Selfridge (1973). This sequence begins 0,0,0,1,1,1,1,... = 0 (3 times), 1 (12 times), 2 (48 times), 3 (192 times), 4 (768 times), ..., and grows too slowly to have its own entry. It is related to the game studied by Hales and Jewett (1963). - N. J. A. Sloane, Dec 02 2016

a(n) are also the incrementally largest terms in the continued fraction of log(2) = [a_0; a_1, a_2, ...] = [0; 1, 2, 3, 1, 6, 3, 1, 1, 2] (excluding the a_0 term). - Eric W. Weisstein, Aug 20 2013

a(n) are also the positions of incrementally highest terms in the continued fraction of log(2) = [a_0; a_1, a_2, ...] = [0; 1, 2, 3, 1, 6, 3, 1, 1, 2] (excluding the term a_0) - Eric W. Weisstein, Aug 20 2013

From Michel Dekking, Aug 11 2022: (Start)

By definition, (a(n)) is an inhomogeneous Beatty sequence. The associated Sturmian word is s(alpha, rho) = (floor((n + 1)*alpha + rho) - floor(n*alpha + rho), n= 0, 1, 2,...) = 1,0,1,1,1,0,1,1,0,1,1,..., with slope alpha = sqrt(2)/2, and intercept rho = -1/2.

Also, s(alpha, rho) = s(alpha,rho+1) - 1. Since 0 < alpha < 1 and 0 < rho +1 < 1, with algebraic conjugates

alpha* = -sqrt(2)/2, and (rho +1)* = 1/2,

Yasutomi's criterion gives that s(alpha, rho) is fixed point of a morphism.

The morphism can be found following the ideas of Chapter 2 in Lothaire's book and Section 4 of my paper "Substitution invariant Sturmian words and binary trees" (cf. A006340).

For a better fit with the literature we will determine the morphism that fixes the binary complement s(1-alpha, 1-(1+rho) ) = 0,1,0,0,0,1,0,0,1,0,0....

Let psi_1 and psi_8 be the elementary Sturmian morphisms given by

psi_1(0)=01 , psi_1(1)=0; psi_8(0)=01, psi(8)1=1.

Let psi := psi_1 psi_8. Then psi is given by

psi(0)=010 , psi(1)=0.

We see that psi fixes the Octanacci sequence A324772.

That psi is the right morphism can be proved by checking that (x,y) = (1-alpha, -rho) is fixed point of the composition T_1 T_8 of the fractional linear maps

T_1(x,y) = ((1-x)/(2-x), (1-y)/(2-x)),

T_8(x,y) = ((1/(2-x), y/(2-x))).

Conclusion, taking the binary complement of psi: the Sturmian word equals A104521. (End)