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because I am trying to find the answer in wolfarmalpa but it show " no solution exist" !

$$a_{n}=a^{3}_{n-1}a^{2}_{n-2} ,\ a_{0}=2 , a_{1}=2a$$

I wrote it like $a(n)=a(n-1)^3*a(n-2)^2,a(0)=2,a(1)=2$

and $a(n)=(a(n-1))^3(a(n-2))^2,a(0)=2,a(1)=2 $

Alex Ravsky
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NANA
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    Which recurrence relation were you trying to find a solution for? – Arthur Apr 16 '18 at 12:44
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    this one $$a_{n}=a^{3}{n-1}a^{2}{n-2} ,\ a_{0}=2 , a_{1}=2a$$ – NANA Apr 16 '18 at 12:49
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    There is no "solution" to be found. It is a definition, Wolfram just does not know a general term for that sequence. – Bill O'Haran Apr 16 '18 at 12:51
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    "WA does not return an answer"$\neq$"We cannot find the answer". Set $a_n=\exp(b_n)$ or $a_n =a\cdot 2^{b_n}$ and wonder if you are able to solve $b_n = 3b_{n-1}+2b_{n-2}$. – Jack D'Aurizio Apr 16 '18 at 12:51
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    Of course it has a solution! You can just start from the given $a_0, a_1$ and generate the solution, term by term. It begins $$a_0 = 1, a_1 = 2a, a_3 = 8a^3, a_4 = 2048a^{11},\ldots$$Whether that solution has some nice closed form expression is a completely different question, and not always one which WolframAlpha is suited to answer. – Arthur Apr 16 '18 at 12:51
  • Thanks all will try to solve it ! – NANA Apr 16 '18 at 12:55
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    @Arthur It should be $a_{\color{red}2}=2^5a^3$ and $a_{\color{red}3}=2^{17}a^{11}$. – Alex Ravsky Nov 26 '20 at 22:38

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By induction it is easy to see that for each non-negative integer $n$ we have $a_n=2^{b_n}a^{c_n}$, where $(b_n)$ and $(c_n)$ are sequences of integer numbers such that $b_0=b_1=1$, $c_0=0$, $c_1=1$ and $b_n=3b_{n-1}+2b_{n-2}$, $c_n=3c_{n-1}+2c_{n-2}$ for each $n\ge 2$. These are linear recurrence with constant coefficients. Their characteristic polynomial $\lambda^2=3\lambda+2$ has distinct roots $\lambda_{1,2}=\tfrac 12\left(9\pm\sqrt{17}\right)$, so it has a general solution $C_1\lambda_1^n+ C_2\lambda_2^n$ for some constants $C_1$ and $C_2$. These coefficients can be easily found from the respective system of linear equations. Namely, for the sequence $(b_n)$ we have $C_1+ C_2=b_0=1$ and $C_1\lambda_1+ C_2\lambda_2=b_1=1$. For the sequence $(b_n)$ we have $C_1+ C_2=c_0=0$ and $C_1\lambda_1+ C_2\lambda_2=c_1=1$.

Alex Ravsky
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