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%I #26 Jan 29 2021 20:46:51
%S 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,
%T 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,
%U 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
%N 2-adic order of the tribonacci sequence.
%H Alois P. Heinz, <a href="/A248174/b248174.txt">Table of n, a(n) for n = 1..10000</a>
%H Diego Marques and Tamás Lengyel, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL17/Lengyel/lengyel21.html">The 2-adic order of the Tribonacci numbers and the equation T_n = m!</a>, Journal of Integer Sequences, Vol. 17 (2014), Article 14.10.1.
%F a(n) = A007814(A000073(n+1)). - _Michel Marcus_, Oct 03 2014
%F From _Amiram Eldar_, Jan 29 2021: (Start)
%F The following 7 formulas completely specify the sequence (Marques and Lengyel, 2014):
%F 1. a(n) = 0 if n == 1 (mod 4) or n == 2 (mod 4).
%F 2. a(n) = 1 if n == 3 (mod 16) or n == 11 (mod 16).
%F 3. a(n) = 2 if n == 4 (mod 16) or n == 8 (mod 16).
%F 4. a(n) = 3 if n == 7 (mod 16).
%F 5. a(n) = A007814(n) - 1 if n == 0 (mod 16).
%F 6. a(n) = A007814(n+4) - 1 if n == 12 (mod 16).
%F 7. a(n) = A007814((n+1)*(n+17)) - 3 if n == 15 (mod 16).
%F Asymptotic mean: lim_{n->oo} (1/n) * Sum_{k=1..n} a(k) = 3/2. (End)
%e For n = 7 we have T_7 = A000073(8) = 24 and the highest power of 2 dividing T_7 is 8 = 2^3.
%p b:= n-> (<<0|1|0>, <0|0|1>, <1|1|1>>^n. <<0, 1, 1>>)[1, 1]:
%p a:= n-> padic[ordp](b(n), 2):
%p seq(a(n), n=1..120); # _Alois P. Heinz_, Oct 03 2014
%t IntegerExponent[LinearRecurrence[{1, 1, 1}, {1, 1, 2}, 100], 2] (* _Amiram Eldar_, Jan 29 2021 *)
%Y Cf. A000073, A007814.
%K nonn
%O 1,4
%A _Jeffrey Shallit_, Oct 03 2014
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