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

Author

Jeffrey Shallit, Oct 03 2014

Keywords

• nonn

Examples

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

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.

Crossrefs

Cf. A000073, A007814.

Programs

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 *)

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)