A093476 Index of occurrence of the first 0 bit in binary representation of 3^n.

2, 3, 2, 5, 2, 2, 3, 2, 4, 2, 2, 3, 2, 3, 2, 5, 2, 2, 3, 2, 4, 2, 2, 3, 2, 3, 2, 6, 2, 2, 3, 2, 4, 2, 2, 3, 2, 4, 2, 7, 2, 2, 3, 2, 5, 2, 2, 3, 2, 4, 2, 2, 3, 2, 3, 2, 5, 2, 2, 3, 2, 4, 2, 2, 3, 2, 3, 2, 5, 2, 2, 3, 2, 4, 2, 2, 3, 2, 3, 2, 6, 2, 2, 3, 2, 4, 2, 2, 3, 2, 4, 2, 7, 2, 2, 3, 2, 5, 2, 2, 3, 2, 4, 2, 2

Offset: 2

Author

Benoit Cloitre, May 22 2004

Keywords

• nonn

Examples

			In binary, 3^5 = [1, 1, 1, 1, 0, 0, 1, 1] where the first 0 occurs at 5th place. Hence a(5)=5.
		

Links

• Robert Israel, Table of n, a(n) for n = 2..10000

Crossrefs

Cf. A011754, A048651.

Programs

Maple

seq(ListTools:-Search(0, ListTools:-Reverse(convert(3^n,base,2))), n=2..200); # Robert Israel, Nov 20 2017

Mathematica

Array[FirstPosition[IntegerDigits[3^#, 2], 0][[1]] &, 105, 2] (* Michael De Vlieger, Nov 20 2017 *)

PARI

a(n)=if(n<2,0,s=1;while(component(binary(3^n),s)>0,s++);s)

Formula

It seems that Sum_{i=2..n} a(i) is asymptotic to c*n with c=2.7(8).....

From Robert Israel, Nov 20 2017: (Start)

a(n) = k if log_2(2 - 1/2^(k-2)) < frac(n*log_2(3)) < log_2(2 - 1/2^(k-1)). By the equidistribution theorem, this occurs with asymptotic density log_2(2-1/2^(k-1)) - log_2(2-1/2^(k-2)).

Thus c = Sum_{k>=2} k (log_2(2-1/2^(k-1)) - log_2(2 - 1/2^(k-2))) = 2 - Sum_{k>=2} log_2(1-1/2^k) = 2.791916824662... Note that A048651 is the decimal expansion of 2^(1-c). (End)