Let a be a fixed integer, and let the sequence H 0 , H 1 , H 2 , . . . be define

ID: 3109270 • Letter: L

Question

Let a be a fixed integer, and let the sequence H0, H1, H2, . . . be defined recursively by H0 = 0, H1 = 1, and Hn+1 = Hn Hn-1 + a for n = 1, 2, ...

Determine whether or not there exists a positive integer k such that Hk = H0 and Hk+1 = H1.

Does the answer depend on a?

(Hint: First set a = 0 and explore the corresponding sequence, see if the sequence becomes repetitive, then set a = 1 and explore, etc. After you gain some experience by exploring particular cases, consider the general case with a any integer. )

Explanation / Answer

H0= 0 and H1= 1

We start recursion from H2

H2 = H1 - H0 + a = 1 - 0 + a = 1 + a

H3 = H2 - H1 + a = 1 + a - 1 + a = 2a

H4 = H3 - H2 + a = 2a - 1 - a + a = 2a - 1

H5 = H4 - H3 + a = 2a - 1 - 2a + a = a - 1

H6 = H5 - H4 + a = a - 1 - 2a + 1 + a = 0

H7 = H6 - H5 + a = 0 - a +1 + a = 1

H8 = H7 - H6 + a = 1 - 0 + a = 1 + a

H9 = H8 - H7 + a = 1 + a - 1 + a = 2a

As we can see that H6 = H0, H7 = H1and H8 = H2. Thus the Hn sequence repeats with period k= 6 with a any integer.

Since the sequence does not depend explicitly on n , the entire Hn sequence is periodic with period k=6.

Thus Hn = Hn + 6 .

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Let a be a fixed integer, and let the sequence H 0 , H 1 , H 2 , . . . be define

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Explanation / Answer

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