Subsequences of the Fibonacci Sequence

Transcription

1 Subsequences of the Fibonacci Sequence William H. Richardson Wichita State University 1 The Fibonacci and Lucas Numbers In this note, we will develop a collection of sequences each of which is a subsequence of the Fibonacci sequence. Each of these sequences has the property that the quotient of consecutive terms converges to a power of the golden ratio. We will begin with a review of the Fibonacci sequence and some of its properties as well as examine a the sequence that yields the Lucas numbers. The sequences we will define involve Fibonacci and Lucas numbers in their definitions. The n th term of the Fibonacci sequence will be denoted by F n. The sequence itself is given by F n+1 = F n + F n 1, with F 0 = F 1 = 1. The first few terms of the Fibonacci sequence are then 1, 1,, 3,, 8, 13, 1, 34,, 89, 144, 33, 377, 610, 987, 197, 84, 4181,.... If we compute the quotients of consecutive terms, we get 1 1, 1, 3, 3, 8, 13 8, 1 13, 34 1, 34, 89, , , , , , , , ,... The decimal representation of these quotients appear to hover around the value The quotient = As it turns out, among the many results that have been discovered about the Fibonacci sequence over the years is that 1

2 F n+1 lim = 1 + n F n =, the golden ratio. This limit can be obtained using the result called the Binet formula 1 ( F n = 1 If we let = ) n+1 ( 1 ) n+1, n = 0, 1,,.... and = 1, we can then write F n = 1 [ n+1 n+1], n = 0, 1,,.... The French mathematician Franois Édouard Anatole Lucas ( ) made a detailed study of the Fibonacci sequence and related sequences. A sequence of the form f n+1 = f n + f n 1 where f 0 and f 1 are given is referred to as a Fibonacci-type sequence. Lucas came up with his own Fibonacci-type sequence given by L n+1 = L n + L n 1, with L 0 = and L 1 = 1. This yields the sequence of numbers, 1, 3, 4, 7, 11, 18, 9, 47,.... One might reasonably ask why start with, 1 instead of 1, 3. If the sequence starts with, 1, then the de Moivre-Binet formula is very nice ( 1 + ) n ( 1 ) n L n = + = n + n, n = 0, 1,,.... In addition one can define L k = ( 1) k L k, for k = 1,, 3,... if the sequence L k starts, 1. 1 There appears to be some controversy as to the credit given for this formula. Most refer to it as the Binet formula in honor of J. P. M. Binet ( ) a French mathematician who published the formula in However, some say Abraham de Moivre ( ) obtained the result one hundred years earlier in 1730, so we will call it the de Moivre-Binet formula.

3 Subsequences of the Fibonacci Sequence We are going to use the Fibonacci numbers and the Lucas numbers to define a collection of sequences as follows x k,n = L k x k,n 1 + ( 1) k 1 x k,n with x k,0 = F 0 and x k,1 = F k, k 1, n. The sequence x 1,n = L 1 x 1,n 1 + ( 1) 0 x 1,n 1 = x 1,n 1 + x 1,n with x 1,0 = 1, x 1,1 = 1, is merely the Fibonacci sequence. The following table gives nineteen terms for the first four of our sequences. n x 1,n x,n x 3,n x 4,n In this table, the second column is the Fibonacci sequence, the third column appears to be every other term of the Fibonacci sequence, the fourth column appears to be every third term of the Fibonacci sequence and the fifth column appears to have every fourth term of the Fibonacci sequence. Moreover, x,18 is very close to the square of the golden ratio, x 3,18 is very close to the x,17 x 3,17 3

4 cube of the golden ratio, and x 4,18 golden ratio. x 4,17 is very close to the fourth power of the We will establish that what we see from the table is in fact true. That is, x k,n generates every k th term of the Fibonacci sequence and that as n goes to x k,n+1 x k,n k for k = 1,, 3,.... We begin with some lemmas. Lemma.1. L k + ( 1)k 1 4 = Fk 1. Proof. In the following, we use the fact that = 1. L k + ( 1) k 1 = ( k + k ) + ( 1) k 1 4 = k + k k + k + ( 1) k 1 4 = k + ( 1) k + k + ( 1) k 1 4 = k ( 1) k + k = ( k k ) = Fk 1 Lemma.. L k + F k 1 = k and L k F k 1 = k Proof. This is just a matter of writing out what the terms are. L k + F k 1 = k + k + k k = k L k F k 1 = k + k k + k = k Lemma.3. F k k F k 1 =. Proof. We begin by calculating F k 1 + k. Again we will use the fact that = 1. 4

5 [ ] 1 F k 1 + k = ( k k ) + k = 1 k+1 1 k + k = 1 k k 1 + k = 1 k k 1 (1 + ) = 1 k k 1 ( ) = 1 ( k+1 k+1 ) = F k. Therefore, solving for we have F k k F k 1 =. We are now ready to prove the major theorem of this note. Theorem.1. x k,n = F nk. Proof. For the recurrence relation x k,n = L k x k,n 1 + ( 1) k 1 x k,n with x k,0 = 1, x k,1 + F k, we have the auxiliary equation r L k r ( 1) k 1 = 0 and r = L k ± L k + ( 1)k 1 4. From Lemma.1, this becomes r = L k ± F k 1. Then an application of Lemma., yields r 1 = k and r = k. Hence x k,n = A( k ) n + B( k ) n with x k,0 = 1, x k,1 = F k.

6 Using the initial conditions that x k,0 = 1 and x k,1 = F k we obtain, A + B = 1 (1) k A + k B = F k () Multiplying equation (1) by k and subtracting from equation () gives ( k k )A = F k K A = F k k k k = F k k Fk 1 An application of Lemma.3 reduces this to A = 1 and then B = 1. Thus, x k,n = 1 ( k ) n 1 ( k ) n = 1 ( nk+1 nk+1 ) = F nk. Theorem.. lim n x k,n+1 x k,n = k. Proof. We note that since > 1 and < 1, it follows that < 1. Therefore, x k,n+1 x k,n x k,n+1 lim n x k,n = F (n+1)k F nk = (n+1)k+1 (n+1)k+1 nk+1 nk+1 = = (n+1)k+1 (n+1)k+1 nk+1 nk+1 1 k 1 k = lim n = k 1 6

7 3 An Interesting Identity In searching for a closed form formula for x k,n two different methods of solving the recurrence relation were used and one ended up with a more complex closed form. However, putting the two solutions together has resulted in an interesting identity. The following theorem gives an alternate method of solving a recurrence relation. Theorem 3.1. Let x n+1 = ax n + bx n 1 with x 0 and x 1 given. If r 1 and r are distinct roots of r ar b = 0, then x n = (x 1 r 1 x 0 )r n (x 1 r x 0 )r n 1 r r 1 (3) Proof. If r 1 and r are the distinct roots of r ar b = 0, then a = r 1 +r and b = r 1 r. We can then write x n+1 ax n = bx n 1 as x n+1 (r 1 + r )x n = r 1 r x n 1. Thus x n+1 r 1 x n = r (x n r 1 x n 1 ) (4) and x n+1 r x n = r 1 (x n r x n 1 ) () Applying the recurrence relation (4) to itself yields x n+1 r 1 x n = r n (x 1 r 1 x 0 ) (6) In a similar manner () gives x n+1 r x n = r1 n (x 1 r x 0 ) (7) Now subtract (6) from (7) to obtain (r 1 r )x n = r1 n (x 1 r x 0 ) r n (x 1 r 1 x 0 ) We only need the result for distinct roots. 7

8 which, in turn, yields If we apply this lemma to x n = (x 1 r x 0 )r n 1 (x 1 r 1 x 0 )r n r 1 r. x k,n = L k x k,n 1 + ( 1) k 1 x k,n with x k,0 = 1, x k,1 = F k, we have, recalling that r 1 = k and r = k, x k,n = (F k k ) nk (F k k ) nk k k = F k nk k nk F k nk + k nk Fk 1 = F k( nk nk ) k k ( (n 1)k (n 1)k ) Fk 1 = Fk F nk 1 ( 1) k F (n 1)k 1 Fk 1 = F kf nk 1 + ( 1) k 1 F (n 1)k 1 F k 1 However in Theorem.1 we found x k,n = F nk. Therefore equating the two forms for x k,n, we have F nk F k 1 = F k F nk 1 + ( 1) k 1 F (n 1)k 1 with n, k 1. 8