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%I #53 Mar 01 2024 06:23:02
%S 1,2,1,2,1,2,2,1,2,1,2,2,1,2,1,2,1,2,2,1,2,1,2,2,1,2,1,2,1,2,2,1,2,1,
%T 2,2,1,2,1,2,1,2,2,1,2,1,2,2,1,2,1,2,2,1,2,1,2,1,2,2,1,2,1,2,2,1,2,1,
%U 2,1,2,2,1,2,1,2,2,1,2,1,2,1,2,2,1,2,1,2,2,1,2,1,2,1,2,2,1,2,1
%N Number of integers m such that 3^n < 2^m < 3^(n+1).
%C Represents increments between successive terms of allowable dropping times in the Collatz (3x+1) problem. That is, a(n) = A020914(n+1) - A020914(n). - _K. Spage_, Oct 23 2009
%H T. D. Noe, <a href="/A022921/b022921.txt">Table of n, a(n) for n = 0..1000</a>
%H Mike Winkler, <a href="https://arxiv.org/abs/1709.03385">The algorithmic structure of the finite stopping time behavior of the 3x + 1 function</a>, arXiv:1709.03385 [math.GM], 2017.
%F a(n) = floor((n+1)*log_2(3)) - floor(n*log_2(3)).
%F a(n) = A122437(n+2) - A122437(n+1) - 1. - _K. Spage_, Oct 23 2009
%F First differences of A020914. - _Robert G. Wilson v_, May 25 2014
%F First differences of A056576. - _L. Edson Jeffery_, Dec 12 2014
%F Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=1..m} a(k) = log_2(3) (A020857). - _Amiram Eldar_, Mar 01 2024
%e From _Amiram Eldar_, Mar 01 2024: (Start)
%e a(0) = 1 because 3^0 = 1 < 2^1 = 2 < 3^1 = 3.
%e a(1) = 2 because 3^1 = 3 < 2^2 = 4 < 2^3 = 8 < 3^2 = 9.
%e a(2) = 1 because 3^2 = 9 < 2^4 = 16 < 3^3 = 27. (End)
%p Digits := 100: c1 := log(3.)/log(2.): A022921 := n->floor((n+1)*c1)-floor(n*c1);
%p seq(ilog2(3^(n+1)) - ilog2(3^n), n=0 .. 1000); # _Robert Israel_, Dec 11 2014
%t i2 = 1; Table[p = i2; While[i2++; 2^i2 < 3^(n + 1)]; i2 - p, {n, 0, 98}] (* _T. D. Noe_, Feb 28 2014 *)
%t f[n_] := Floor[ Log2[ 3^n] + 1]; Differences@ Array[f, 106, 0] (* _Robert G. Wilson v_, May 25 2014 *)
%o (PARI) a(n) = logint(3^(n+1),2) - logint(3^n,2) \\ _Ruud H.G. van Tol_, Dec 28 2022
%o (PARI) Vec(matreduce([logint(2^i,3)|i<-[1..158]])[,2])[1..-2] \\ _Ruud H.G. van Tol_, Dec 29 2022
%Y Cf. A020914, A056576, A076227, A100982, A122437.
%Y See also A020857 (decimal expansion of log_2(3)).
%K nonn,easy
%O 0,2
%A _Clark Kimberling_
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