A271591 Second most significant bit of the tribonacci number A000073(n).

0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1

Offset: 4

Author

Andres Cicuttin, Apr 10 2016

Keywords

• nonn
• base

Comments

It is conjectured that after the first two 0's, the number of consecutive 0's is only 4 or 5, and the number of consecutive 1's is only 3 or 4 (tested up to n=10^4). The sequence looks quasiperiodic (or with a very long true period if any).

Examples

			(Second MSB in parenthesis)
  n   A000073(n)      A000073(n)
      decimal         binary
  4      2        ->   1(0)
  5      4        ->   1(0)0
  6      7        ->   1(1)1
  7      13       ->   1(1)01
  8      24       ->   1(1)000
  9      44       ->   1(0)1100
  10     81       ->   1(0)10001
  11     149      ->   1(0)010101
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 4..10000

Crossrefs

Cf. A000073 (tribonacci numbers), A079944 (2nd msb), A272170.

Programs

Mathematica

a = LinearRecurrence[{1, 1, 1}, {0, 0, 1}, 120];(* to generate A000073 *)
Table[IntegerDigits[a, 2][[i]][[2]], {i, 5, Length[a]}]

Python

A271591_list, a, b, c = [], 0, 1 ,1
for n in range(4,10001):
    a, b, c = b, c, a+b+c
    A271591_list.append(int(bin(c)[3])) # Chai Wah Wu, Feb 07 2018

Formula

a(n) = floor(A000073(n)/(2^(ceiling(log_2(A000073(n) + 1)) - 2))) - 2.

a(n) = A079944(A000073(n)-2). - Michel Marcus, Apr 22 2016