A056576 Highest k with 2^k <= 3^n.

0, 1, 3, 4, 6, 7, 9, 11, 12, 14, 15, 17, 19, 20, 22, 23, 25, 26, 28, 30, 31, 33, 34, 36, 38, 39, 41, 42, 44, 45, 47, 49, 50, 52, 53, 55, 57, 58, 60, 61, 63, 64, 66, 68, 69, 71, 72, 74, 76, 77, 79, 80, 82, 84, 85, 87, 88, 90, 91, 93, 95, 96, 98, 99, 101, 103, 104, 106, 107

Offset: 0

Author

Henry Bottomley, Jun 29 2000

Keywords

• nonn

Examples

			a(3)=4 because 3^3=27 and 2^4=16 is power of 2 immediately below 27.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..10000
• Mike Winkler, The algorithmic structure of the finite stopping time behavior of the 3x+ 1 function, arXiv:1709.03385 [math.GM], 2017.

Crossrefs

Cf. A000079 (powers of 2), A000244 (powers of 3), A020914, A022921.

Cf. A056850, A117630 (complement), A020857 (decimal expansion of log_2(3)), A076227, A100982.

Programs

Haskell

a056576 = subtract 1 . a020914  -- Reinhard Zumkeller, May 17 2015

Maple

seq(ilog2(3^n), n= 0 .. 1000); # Robert Israel, Dec 11 2014

Mathematica

Table[Floor[Log[2, 3^n]], {n, 0, 69}] (* Robert G. Wilson v, Apr 06 2006 *)
Table[Floor[n*Log[2, 3]], {n, 0, 68}] (* L. Edson Jeffery, Dec 11 2014 *)

PARI

{a(n) = if( n<0, 0, logint(3^n, 2))}; /* Michael Somos, Dec 13 2014 */

Python

def A056576(n): return (3**n).bit_length()-1 # Chai Wah Wu, Oct 09 2024

Formula

a(n) = floor(log_2(3^n)) = log_2(A000244(n)-A056576(n)) = a(n-1)+A022921(n-1).

a(n) = A020914(n) - 1. - L. Edson Jeffery, Dec 12 2014