Special sequence

Algebra Level 4

A function J ( x ) J(x) follows the following conditions:

{ J ( 0 ) = J ( 1 ) = 2 J ( 2 ) = J ( 1 ) + J ( 0 ) + 2 = 6 J ( 3 ) = J ( 1 ) + 1 = 3 J ( x ) = J ( ∣ x − 1 ∣ ) + J ( ∣ x − 2 ∣ ) + J ( ∣ x − 3 ∣ ) − J ( ∣ x − 4 ∣ ) ; x > 3 \begin{cases} J(0) = J(1) = 2 \\ J(2) = J(1) + J(0) + 2 = 6 \\ J(3) = J(1) + 1 = 3 \\ J(x) = J(|x-1|) + J(|x-2|) + J(|x-3|) - J(|x-4|); \quad x > 3 \end{cases}

Then, find the value of J ( 9 ) J(9) .


The answer is 123.

Viki Zeta by Viki Zeta , 19, India

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1 solution

Roger Erisman
Aug 19, 2016

For x > 3 the function J(x) is simply the sum of the 3 immediately preceding function values minus the value that is 4 preceding.

The values of the function for x = 4, 5, 6, 7 ,8, and 9 are therefore 9,16,22,44,73, and 123.

J(4) = 3 + 6 +2 - 2

J(5) = 9 +3 + 6 -2

J(6) = 16 + 9 + 3 - 6

J(7) = 22 + 16 + 9 - 3

J(8) = 44 + 22 + 16 - 9

J(9) = 73 + 44 + 22 - 16

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