A248174 2-adic order of the tribonacci sequence.

0, 0, 1, 2, 0, 0, 3, 2, 0, 0, 1, 3, 0, 0, 6, 3, 0, 0, 1, 2, 0, 0, 3, 2, 0, 0, 1, 4, 0, 0, 6, 4, 0, 0, 1, 2, 0, 0, 3, 2, 0, 0, 1, 3, 0, 0, 7, 3, 0, 0, 1, 2, 0, 0, 3, 2, 0, 0, 1, 5, 0, 0, 7, 5, 0, 0, 1, 2, 0, 0, 3, 2, 0, 0, 1, 3, 0, 0, 6, 3, 0, 0, 1, 2, 0, 0, 3, 2, 0, 0, 1, 4, 0, 0, 6, 4, 0, 0, 1, 2, 0, 0, 3, 2

Offset

1,4

Links

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Diego Marques and Tamás Lengyel, The 2-adic order of the Tribonacci numbers and the equation T_n = m!, Journal of Integer Sequences, Vol. 17 (2014), Article 14.10.1.

Formula

a(n) = A007814(A000073(n+1)). - Michel Marcus, Oct 03 2014

From Amiram Eldar, Jan 29 2021: (Start)

The following 7 formulas completely specify the sequence (Marques and Lengyel, 2014):

1. a(n) = 0 if n == 1 (mod 4) or n == 2 (mod 4).

2. a(n) = 1 if n == 3 (mod 16) or n == 11 (mod 16).

3. a(n) = 2 if n == 4 (mod 16) or n == 8 (mod 16).

4. a(n) = 3 if n == 7 (mod 16).

5. a(n) = A007814(n) - 1 if n == 0 (mod 16).

6. a(n) = A007814(n+4) - 1 if n == 12 (mod 16).

7. a(n) = A007814((n+1)*(n+17)) - 3 if n == 15 (mod 16).

Asymptotic mean: lim_{n->oo} (1/n) * Sum_{k=1..n} a(k) = 3/2. (End)

Example

For n = 7 we have T_7 = A000073(8) = 24 and the highest power of 2 dividing T_7 is 8 = 2^3.

Maple

b:= n-> (<<0|1|0>, <0|0|1>, <1|1|1>>^n. <<0, 1, 1>>)[1, 1]:
a:= n-> padic[ordp](b(n), 2):
seq(a(n), n=1..120);  # Alois P. Heinz, Oct 03 2014

Mathematica

IntegerExponent[LinearRecurrence[{1, 1, 1}, {1, 1, 2}, 100], 2] (* Amiram Eldar, Jan 29 2021 *)

Crossrefs

Cf. A000073, A007814.

Sequence in context: A319668 A167634 A174169 * A125095 A113411 A260110

Adjacent sequences: A248171 A248172 A248173 * A248175 A248176 A248177

Keyword

nonn

Author

Jeffrey Shallit, Oct 03 2014

Status

approved