Loog at this Familiar Recursive!

$a_1= 3^0, a_2=3^1, a_n=a_{n-1} a_{n-2}$

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Thus if we define $b_n = \log_3 (a_n)$ then we get a recurrence

$b_n = b_{n-1}+b_{n-2}$ with start conditions $b_1=0$ , $b_2=1$

Thus we get $b_n = F_{n-1}$ where $F_n$ is $n^{th}$ term of the fibonacci sequence.

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Hence $b_{514} = F_{513} \equiv 8 \pmod{10}$ (we count this in $\pmod{10}$ to apply Fermat's little theorem later.

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$a_n = 3^{b_n}$ as defined,

$a_{514} = 3^{F_{513}} \equiv 3^{10k+8} \pmod{11}$ (Because of the fact seen above, $F_{513} \equiv 8 \pmod{10}$

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And now by Fermat's little theorem,

$a_{514}= 3^{F_{513}} \equiv 3^8 \equiv 5 \pmod{11}$

Hence answer is $\boxed{5}$

The given series can be written as 3^0, 3^1, 3^1, 3^2, 3^3, 3^5...=> 3 to the power of fibonacci series from second term. 3^5 leaves remainder of 1 when divided by 11. Every fifth Fibonacci number is the multiple of five. 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 and so on. First term, sixth term, eleventh term etc. of the given series leaves the remainder of 1... Divide the Fibonacci series by 5 and write remainder alone. 1 1 2 3 0 3 3 1 4 0 4 4 3 2 0 2 2 4 1 0 1 1 2 3 0 . This remainder series start recuring from 20th term. Hence 513/20 leaves 13 as remainder. The thirteenth term in that remainder series (which is obtained by dividing Fibonacci series by 5) is 3. Hence 3^3/11 => 27/11 leaves 5 as remainder .

I simply wrote down the recursion .I was lucky enough to find a repitition of the residues mod 11 after 20 terms.