2016th term of the sequence

a m + n + a m n = 1 2 ( a 2 m + a 2 n ) a_{m+n} + a_{m-n} = \frac{1}{2}(a_{2m}+a_{2n})

The sequence a 0 , a 1 , a 2 , . . . a_0,a_1,a_2,... satisfies the above relation for all non-negative integers m m and n n with m n m \geq n .

If a 1 = 1 a_1 = 1 , determine a 2016 a_{2016} .


The answer is 4064256.

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1 solution

Aryaman Maithani
Jul 10, 2018

For the sake of convenience, I'm rewriting the notation of a n a_n as f ( n ) f(n) .

Given: f ( m + n ) + f ( m n ) = 1 2 ( f ( 2 m ) + f ( 2 n ) ) f(m+n) + f(m-n) = \dfrac{1}{2}(f(2m)+f(2n))

Let m + n = X , m n = Y m+n = X, m-n = Y

2 m = X + Y , 2 n = X Y \therefore 2m = X+Y, 2n = X-Y

f ( X ) + f ( Y ) = 1 2 ( f ( X + Y ) + f ( X Y ) ) \therefore f(X) + f(Y) = \dfrac{1}{2}(f(X+Y) + f(X-Y))


Subsituting X = Y = 0 : X = Y= 0:

2 f ( 0 ) = f ( 0 ) f ( 0 ) = 0 2f(0) = f(0) \implies f(0) = 0


Substituting X = Y = 1 : X=Y=1:

2 f ( 1 ) = 1 2 ( f ( 2 ) + f ( 0 ) ) f ( 2 ) = 4 2f(1) = \dfrac{1}{2}(f(2)+f(0)) \implies f(2) = 4


Subsituting Y = 1 : Y=1:

f ( X ) + f ( 1 ) = 1 2 ( f ( X + 1 ) f ( X 1 ) ) f(X) + f(1) = \dfrac{1}{2}(f(X+1) - f(X-1))

f ( X + 1 ) = 2 f ( X ) f ( X 1 ) + 2 \implies f(X+1) = 2f(X) - f(X-1) + 2

Substituting X = 2 X = 2 gives f ( 3 ) = 9 f(3) = 9

It looks likes the following could be true: f ( X ) = X 2 X N 0 f(X) = X^2 \forall X \in \mathbb{N} \cup {0}

And this can be proved easily via induction, skipping the complete formal proof and just demonstrating the inductive step:

f ( X + 1 ) = 2 f ( X ) f ( X 1 ) + 2 f(X+1) = 2f(X) - f(X-1)+2

f ( X + 1 ) = 2 X 2 ( X 1 ) 2 + 2 \implies f(X+1) = 2X^2 - (X-1)^2 + 2

f ( X + 1 ) = 2 X 2 X 2 + 2 X 1 + 2 = X 2 + 2 X + 1 = ( X + 1 ) 2 \implies f(X+1) = 2X^2 - X^2 + 2X -1 +2 = X^2 + 2X +1 = (X+1)^2

f ( X ) = X 2 a 2016 = f ( 2016 ) = 201 6 2 = 4064256 \therefore f(X) = X^2 \implies a_{2016} = f(2016) = 2016^2 = \boxed{4064256}

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