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Beatty sequence for 2^x, where 1/x + 1/2^x = 1.
3

%I #4 Jan 03 2020 20:20:08

%S 2,5,8,11,14,17,20,23,25,28,31,34,37,40,43,46,49,51,54,57,60,63,66,69,

%T 72,75,77,80,83,86,89,92,95,98,101,103,106,109,112,115,118,121,124,

%U 127,129,132,135,138,141,144,147,150,153,155,158,161,164,167,170

%N Beatty sequence for 2^x, where 1/x + 1/2^x = 1.

%C Let x be the solution of 1/x + 1/2^x = 1. Then (floor(n x)) and (floor(n 2^x)) are a pair of Beatty sequences; i.e., every positive integer is in exactly one of the sequences. See the Guide to related sequences at A329825.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/BeattySequence.html">Beatty Sequence.</a>

%H <a href="/index/Be#Beatty">Index entries for sequences related to Beatty sequences</a>

%F a(n) = floor(n 2^x), where x = 1.52980838275... is the constant in A329986.

%t r = x /. FindRoot[1/x + 1/2^x == 1, {x, 1, 10}, WorkingPrecision -> 120]

%t RealDigits[r][[1]] (* A329986 *)

%t Table[Floor[n*r], {n, 1, 250}] (* A329987 *)

%t Table[Floor[n*2^r], {n, 1, 250}] (* A329988 *)

%Y Cf. A329825, A329986, A329987 (complement).

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Jan 02 2020