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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
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).
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: A374220 A335716 A138150 * A287790 A285159 A073089
KEYWORD
nonn,base
AUTHOR
Andres Cicuttin, Apr 10 2016
STATUS
approved