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A271591 Second most significant bit of the tribonacci number A000073(n). 4
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 (list; graph; refs; listen; history; text; internal format)
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: A165560 A014306 A138150 * A287790 A285159 A073089

Adjacent sequences:  A271588 A271589 A271590 * A271592 A271593 A271594

KEYWORD

nonn,base

AUTHOR

Andres Cicuttin, Apr 10 2016

STATUS

approved

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Last modified November 15 01:22 EST 2019. Contains 329142 sequences. (Running on oeis4.)