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

4,1

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).

Links

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

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

Example

(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

Mathematica

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

Prog

(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

Crossrefs

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

Sequence in context: A014306 A335716 A138150 * A287790 A285159 A073089

Adjacent sequences: A271588 A271589 A271590 * A271592 A271593 A271594

Keyword

nonn,base

Author

Andres Cicuttin, Apr 10 2016

Status

approved