A022921 Number of integers m such that 3^n < 2^m < 3^(n+1).
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, 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, 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
Offset: 0
Examples
From _Amiram Eldar_, Mar 01 2024: (Start) a(0) = 1 because 3^0 = 1 < 2^1 = 2 < 3^1 = 3. a(1) = 2 because 3^1 = 3 < 2^2 = 4 < 2^3 = 8 < 3^2 = 9. a(2) = 1 because 3^2 = 9 < 2^4 = 16 < 3^3 = 27. (End)
Links
- T. D. Noe, Table of n, a(n) for n = 0..1000
- Mike Winkler, The algorithmic structure of the finite stopping time behavior of the 3x + 1 function, arXiv:1709.03385 [math.GM], 2017.
Crossrefs
Programs
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Maple
Digits := 100: c1 := log(3.)/log(2.): A022921 := n->floor((n+1)*c1)-floor(n*c1); seq(ilog2(3^(n+1)) - ilog2(3^n), n=0 .. 1000); # Robert Israel, Dec 11 2014
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Mathematica
i2 = 1; Table[p = i2; While[i2++; 2^i2 < 3^(n + 1)]; i2 - p, {n, 0, 98}] (* T. D. Noe, Feb 28 2014 *) f[n_] := Floor[ Log2[ 3^n] + 1]; Differences@ Array[f, 106, 0] (* Robert G. Wilson v, May 25 2014 *)
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PARI
a(n) = logint(3^(n+1),2) - logint(3^n,2) \\ Ruud H.G. van Tol, Dec 28 2022
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PARI
Vec(matreduce([logint(2^i,3)|i<-[1..158]])[,2])[1..-2] \\ Ruud H.G. van Tol, Dec 29 2022
Formula
a(n) = floor((n+1)*log_2(3)) - floor(n*log_2(3)).
First differences of A020914. - Robert G. Wilson v, May 25 2014
First differences of A056576. - L. Edson Jeffery, Dec 12 2014
Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=1..m} a(k) = log_2(3) (A020857). - Amiram Eldar, Mar 01 2024
Comments