A204399 Numbers k such that floor(2^k / 3^n) = 1.

0, 2, 4, 5, 7, 8, 10, 12, 13, 15, 16, 18, 20, 21, 23, 24, 26, 27, 29, 31, 32, 34, 35, 37, 39, 40, 42, 43, 45, 46, 48, 50, 51, 53, 54, 56, 58, 59, 61, 62, 64, 65, 67, 69, 70, 72, 73, 75, 77, 78, 80, 81, 83, 85, 86, 88, 89, 91, 92, 94, 96, 97, 99, 100, 102, 104

Offset: 0

Author

Michel Lagneau, Jan 15 2012

Keywords

• nonn

Comments

Presumably a(n) ~ 3*n - floor(n*sqrt(2)) = A195176(n). In the first hundred, a(n) = A195176(n) except for n = 41, 70, 82, 94 where a(n) = A195176(n) - 1.

The conjecture is false; A195176(n) - a(n) increases without bound (though not monotonically) since log_2(3) < 3 - sqrt(2). - Charles R Greathouse IV, Jan 15 2012

Basically a duplicate of A020914. - R. J. Mathar, Jan 16 2012

Crossrefs

Cf. A020914, A056576.

Programs

Maple

for n from 0 to 120 do : for k from 0 to 100 do: x:=floor(2^k /3^n):if x=1 then printf(`%d, `,k):else fi:od:od:

PARI

a(n)=ceil(n*log(3)/log(2)) \\ Charles R Greathouse IV, Jan 15 2012

Formula

a(n) = ceiling( n * log_2(3) ). - Charles R Greathouse IV, Jan 15 2012