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A248174
2-adic order of the tribonacci sequence.
2
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
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
Sequence in context: A319668 A167634 A174169 * A125095 A113411 A260110
KEYWORD
nonn
AUTHOR
Jeffrey Shallit, Oct 03 2014
STATUS
approved