TITLE: Prove or disprove that $2^n$ divides $T_{2^n}$ for $n > 2$.
QUESTION [7 upvotes]: The Tribonacci sequence satisfies
$$T_0 = T_1 = 0, T_2 = 1,$$
$$T_n = T_{n-1} + T_{n-2} + T_{n-3}.$$
Prove or disprove that $2^n$ divides $T_{2^n}$ for $n > 2$. 
(I think $2^n$ divides $T_{2^n}$.)
P.S.
To confirm the ArtW's proposition, I calculate $T_{2^n + l}$ mod $2^{n + 2}$ by ruby.
require 'matrix'

def power(a, n, mod)
  return Matrix.I(a.row_size) if n == 0
  m = power(a, n >> 1, mod)
  m = (m * m).map{|i| i % mod}
  return m if n & 1 == 0
  (m * a).map{|i| i % mod}
end

def f(m, n, mod)
  ary0 = Array.new(m, 0)
  ary0[0] = 1
  v = Vector.elements(ary0)
  ary1 = [Array.new(m, 1)]
  (0..m - 2).each{|i|
    ary2 = Array.new(m, 0)
    ary2[i] = 1
    ary1 << ary2
  }
  a = Matrix[*ary1]
  (power(a, n, mod) * v)[m - 1]
end

[-2, -1, 0, 1, 2].each{|l|
  (1..20).each{|i|
    j = 2 ** i + l
    # T_j % (2 ** (i + 2))
    p [j, f(3, j, 2 ** (i + 2))]
  }
}

Output
[0, 0]
[2, 1]
[6, 7]
[14, 31]
[30, 127]
[62, 255]
[126, 511]
[254, 1023]
[510, 2047]
[1022, 4095]
[2046, 8191]
[4094, 16383]
[8190, 32767]
[16382, 65535]
[32766, 131071]
[65534, 262143]
[131070, 524287]
[262142, 1048575]
[524286, 2097151]
[1048574, 4194303]
[1, 0]
[3, 1]
[7, 13]
[15, 41]
[31, 17]
[63, 33]
[127, 65]
[255, 129]
[511, 257]
[1023, 513]
[2047, 1025]
[4095, 2049]
[8191, 4097]
[16383, 8193]
[32767, 16385]
[65535, 32769]
[131071, 65537]
[262143, 131073]
[524287, 262145]
[1048575, 524289]
[2, 1]
[4, 2]
[8, 24]
[16, 0]
[32, 64]
[64, 128]
[128, 256]
[256, 512]
[512, 1024]
[1024, 2048]
[2048, 4096]
[4096, 8192]
[8192, 16384]
[16384, 32768]
[32768, 65536]
[65536, 131072]
[131072, 262144]
[262144, 524288]
[524288, 1048576]
[1048576, 2097152]
[3, 1]
[5, 4]
[9, 12]
[17, 8]
[33, 80]
[65, 160]
[129, 320]
[257, 640]
[513, 1280]
[1025, 2560]
[2049, 5120]
[4097, 10240]
[8193, 20480]
[16385, 40960]
[32769, 81920]
[65537, 163840]
[131073, 327680]
[262145, 655360]
[524289, 1310720]
[1048577, 2621440]
[4, 2]
[6, 7]
[10, 17]
[18, 49]
[34, 33]
[66, 65]
[130, 129]
[258, 257]
[514, 513]
[1026, 1025]
[2050, 2049]
[4098, 4097]
[8194, 8193]
[16386, 16385]
[32770, 32769]
[65538, 65537]
[131074, 131073]
[262146, 262145]
[524290, 524289]
[1048578, 1048577]

REPLY [3 votes]: A solution by matrix form (as suggested by Ihf):
We can write:
$\left(\begin{array}{c}
T_{n+1}\\
T_{n+2}\\
T_{n+3}
\end{array}\right)=\left(\begin{array}{ccc}
0 & 1 & 0\\
0 & 0 & 1\\
1 & 1 & 1
\end{array}\right)\left(\begin{array}{c}
T_{n}\\
T_{n+1}\\
T_{n+2}
\end{array}\right)$
So if $M$ is the matrix of multiplication, we can write $(T_n,T_{n+1},T_{n+2})^t=M^n\cdot(0,0,1)^t$.
In particular $T_n$ is the value in the right-top corner of the matrix $M^n$, so we have to show that for $n>2$ the righ-top corner of the matrix $M^{2^n}$ is divisible by $2^n$. Let's do it by recursion: let $n$ be an integer such that we can write the matrix $M^{2^n} \pmod{2^n}$ in the following form (for example $n=3$):
$\left(\begin{array}{ccc}
2^{n-1}+1 & 2^{n-1} & 0\\
0 & 2^{n-1}+1 & 2^{n-1}\\
2^{n-1} & 2^{n-1} & 1
\end{array}\right)$
Now we lift it to a matrix $\pmod{2^{n+1}}$, clearly we have more possibilities, so we will write it in the following form: 
$\left(\begin{array}{ccc}
2^{n-1}+1+\delta_1\cdot2^n & 2^{n-1}+\delta_2\cdot2^n & \delta_3\cdot2^n\\
\delta_4\cdot2^n & 2^{n-1}+1+\delta_5\cdot2^n & 2^{n-1}+\delta_6\cdot2^n\\
2^{n-1}+\delta_7\cdot2^n & 2^{n-1}+\delta_8\cdot2^n & 1+\delta_9\cdot2^n
\end{array}\right)$
where $\delta_i=0,1$.
If we calculate the square of the matrix (always mod $2^{n+1}$) we found a matrix in the same form of the case $2^n$.