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

%I #7 Jan 24 2020 11:11:03

%S 2,4,7,9,12,14,17,19,22,24,27,29,31,34,36,39,41,44,46,49,51,54,56,59,

%T 61,63,66,68,71,73,76,78,81,83,86,88,90,93,95,98,100,103,105,108,110,

%U 113,115,118,120,122,125,127,130,132,135,137,140,142,145,147

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

%C Let x be the solution of 1/x^2 + 1/2^x = 1. Then (floor(n x^2)) 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 = 1.298192... is the constant in A329992.

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

%t RealDigits[r][[1]] (* A329992 *)

%t Table[Floor[n*r^2], {n, 1, 250}] (* A329993 *)

%t Table[Floor[n*2^r], {n, 1, 250}] (* A329994 *)

%Y Cf. A329825, A329992, A329993 (complement).

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Jan 02 2020

%E Formula corrected. - _R. J. Mathar_, Jan 24 2020