2016 is absolutely awesome 2

a 0 , a 1 , . . . a_0, a_1,... is a sequence of positive integers where a 0 = 1 ; a 1 = 1 ; a 2 = 2 ; a 3 = 6 a_0=1; a_1=1; a_2=2; a_3=6 .

Moreover, for all n 4 n \ge 4 , a n a_n is the smallest positive integer such that:

a n a i a n i \dfrac{a_n}{a_ia_{n-i}} is an integer for all integers i , 0 i n i, 0 \le i \le n .

Find a 2016 2016 \sqrt[2016]{a_{2016}} .


This is a part of the Set .


The answer is 2.040.

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1 solution

Note that it is equivalent to saying that for n 4 n\ge 4 , a n a_n is the least common multiple of { a 1 a n 1 , a 2 a n 2 , , a n 1 a 1 } \{a_1a_{n-1}, a_2a_{n-2}, \ldots,a_{n-1}a_1\} .

This formulation makes it clear that an has no prime factors other than 2 2 and 3 3 , for any n n . We also notice that we can consider these two prime factors independently, since the LCM of the above set will be the highest power of two present times the highest power of three present.

In particular, let b n b_n denote the sequence defined as above, except b 0 = b 1 = 1 b_0 = b_1 = 1 and b 2 = b 3 = 2 b_2 = b_3 = 2 .

Similarly, let c n c_n denote the sequence defined as above, except c 0 = c 1 = c 2 = 1 c_0 = c_1 = c_2 = 1 and c 3 = 3 c_3 = 3 .

It is clear that a n = b n c n a_n = b_nc_n for all n n .

We claim that b n = 2 n / 2 b_n = 2^{\left\lfloor n/2\right\rfloor} . The proof is by induction.

If n = 2 k n = 2k , then b i b 2 k i = 2 i / 2 . 2 ( 2 k i ) / 2 2 i / 2 . 2 k i / 2 = 2 k b_ib_{2k-i} = 2^{\left\lfloor i/2\right\rfloor}.2^{\left\lfloor (2k-i)/2\right\rfloor}\le 2^{i/2}.2^{k-i/2}= 2^k , with equality when i 0 ( m o d 2 ) i \equiv 0 \pmod 2 .

And, if n = 2 k + 1 n = 2k + 1 , then b i b 2 k + 1 i = 2 i / 2 . 2 ( 2 k + 1 i ) / 2 = 2 i / 2 . 2 ( 2 k + 1 i ) / 2 . 2 1 / 2 = 2 k b_ib_{2k+1-i} = 2^{\left\lfloor i/2\right\rfloor}.2^{\left\lfloor (2k+1-i)/2\right\rfloor}=2^{i/2}.2^{(2k+1-i)/2}.2^{-1/2}= 2^k , because exactly one of i i and 2 k + 1 i 2k+1-i is odd.

A similar argument, with 3 cases modulo 3 3 , works to show c n = 3 n / 3 c_n = 3^{\left\lfloor n/3\right\rfloor} .

Hence, in general we have a n = 2 n / 2 3 n / 3 a_n = 2^{\left\lfloor n/2\right\rfloor}3^{\left\lfloor n/3\right\rfloor} .

So we have a 2016 2016 = 2 1008 . 3 672 2016 = 2 . 3 3 2.040 \sqrt[2016]{a_{2016}}=\sqrt[2016]{2^{1008}.3^{672}}=\sqrt2.\sqrt[3]{3}\approx\boxed{2.040} .

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