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A022921 Number of integers m such that 3^n < 2^m < 3^(n+1). 10
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
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
LINKS
FORMULA
a(n) = floor((n+1)*log_2(3)) - floor(n*log_2(3)).
a(n) = A122437(n+2) - A122437(n+1) - 1. - K. Spage, Oct 23 2009
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
EXAMPLE
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)
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
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 *)
PROG
(PARI) a(n) = logint(3^(n+1), 2) - logint(3^n, 2) \\ Ruud H.G. van Tol, Dec 28 2022
(PARI) Vec(matreduce([logint(2^i, 3)|i<-[1..158]])[, 2])[1..-2] \\ Ruud H.G. van Tol, Dec 29 2022
CROSSREFS
See also A020857 (decimal expansion of log_2(3)).
Sequence in context: A326194 A331251 A309858 * A080763 A245920 A165413
KEYWORD
nonn,easy
AUTHOR
STATUS
approved

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Last modified May 5 09:50 EDT 2024. Contains 372269 sequences. (Running on oeis4.)