Prove or disprove that $2^n$ divides $T_{2^n}$ for $n > 2$.

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]


I'll show that $2^{n+1}\mid\mid T_{2^n}$ for all $n\ge 5$. Together with casework for $n<5$ this gives a proof of your claim.

Lemma: For all integers $n,m\ge 1$ we have $$T_{n+m}=T_{m}T_{n-1}+T_{m-1}T_{n}+T_{m}T_{n}+T_{m+1}T_{n+1}.$$ Proof: Straightforward induction on $m$. $\square$

Proposition: For all integers $n\ge 5$: $$T_{2^n}\equiv 2^{n+1}\pmod{2^{n+2}}.$$ Proof: We shall use induction on $n$ to prove simultaneously the following congruences for $n\ge 5$: $$\begin{cases} T_{2^{n}-2}&\equiv -1&&\pmod{2^{n+2}}\\T_{2^{n}-1}&\equiv 2^{n-1}+1&&\pmod{2^{n+2}}\\T_{2^{n}}&\equiv 2^{n+1}&&\pmod{2^{n+2}}\\T_{2^{n}+1}&\equiv 5\cdot 2^{n-1}&&\pmod{2^{n+2}}\\T_{2^{n}+2}&\equiv 2^{n}+1&&\pmod{2^{n+2}}\end{cases}$$ This is true for $n=5$. Assume it to be true for some $n\ge 5$. Then by the lemma, \begin{align}T_{2^{n+1}}&=T_{2^n-1}T_{2^n}+T_{2^n-2}T_{2^n+1}+T_{2^n-1}T_{2^n+1}+T_{2^n}T_{2^n+2}\\&\equiv(2^{n-1}+1)2^{n+1}-5\cdot2^{n-1}+(2^{n-1}+1)\cdot 5\cdot 2^{n-1}+2^{n+1}(2^n+1)\pmod{2^{n+3}}\\&\equiv2^{n+2}\pmod{2^{n+3}}\end{align} For $T_{2^{n+1}-1}$ and $T_{2^{n+1}+1}$ we use something similar, so I let that to you. Finally, $T_{2^{n+1}-2}$ and $T_{2^{n+1}+2}$ are found using the recursion formula. This completes induction. $\square$

• It'd be nice to interpret this in matrix form as in math.stackexchange.com/questions/1827074/…. However, the powers of the corresponding matrix are not so immediate because they mix two Tribonacci sequence with different starting values. – lhf Jun 21 '16 at 13:27
• \begin{align}T_{2^{n+1}}&=T_{2^n-1}T_{2^n}+T_{2^n-2}T_{2^n+1}+T_{2^n-1}T_{2^n+1}+T_{2^n}T_{2^n+2}\\&\equiv(2^{n-1}+1)2^{n+1}-5\cdot2^{n-1}+(2^{n-1}+1)\cdot 5\cdot 2^{n-1}+2^{n+1}(2^n+1)\pmod{2^{n+2}}\\&\equiv?\pmod{2^{n+3}}\end{align} – TOM Jun 25 '16 at 22:18
• @Manyama We can take the congruences mod $2^{n+3}$ because every term is divisible by 2: if $a\equiv b \pmod{2^n}$ then $2a\equiv 2b\pmod{2^{n+1}}$. – ArtW Jun 27 '16 at 23:12
• Is the following OK? \begin{align}T_{2^{n+1}}&=T_{2^n-1}T_{2^n}+T_{2^n-2}T_{2^n+1}+T_{2^n-1}T_{2^n+‌​1}+T_{2^n}T_{2^n+2}\\&\equiv(2^{n-1}+1)2^{n+1}-5\cdot2^{n-1}+(2^{n-1}+1)\cdot 5\cdot 2^{n-1}+2^{n+1}(2^n+1)\pmod{2^{n+2}}\\&\equiv2\cdot(2^{n-1}+1)2^{n+1}-2\cdot 5\cdot2^{n-1}+2\cdot (2^{n-1}+1)\cdot 5\cdot 2^{n-1}+2\cdot 2^{n+1}(2^n+1)\pmod{2^{n+3}}\\&\equiv0\pmod{2^{n+3}}\\&\end{align} – TOM Jun 28 '16 at 10:01
• no, you should factor out a 2 instead of adding a factor! – ArtW Jun 28 '16 at 11:49

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 $2^n$.