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A Universal Convolution Identity for Fibonacci-Type and Lucas-Type Sequences

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A Universal Convolution Identity for Fibonacci-Type and Lucas-Type Sequences A Universal Convolution Identity for Fibonacci-type and Lucas-type Sequences. Greg Dresden Yichen Wang Washington and Lee University University of California, Los Angeles Lexington, VA 24450 Los Angeles, CA 90095 [email protected] [email protected] We will define all these numbers in a moment, but first let us write down some convolution formulas and see if any interesting patterns appear. We start with this delightful and well-known convolution formula for the Fibonacci and Lucas numbers, n X FiLn−i = (n + 1)Fn: (1) i=0 The Pell and Pell-Lucas numbers Pn;Qn from [4] satisfy the somewhat similar equation n−1 X 1 P Q = (n − 1)P : (2) i n−i 2 n i=1 As for the Padovan and Perrin numbers An and En from [7, 9, 11], we have n X AiEn−i = (n + 5)An − En+2: (3) i=0 Finally, in [3, Proposition 1] we find this identity for the the Tribonacci numbers Tn, n−3 X TiUn−i = (n − 2)Tn−1 − Tn−2; (4) i=0 with Un defined as Un = Tn + Tn−2 + 2Tn−3 for n ≥ 3. Although it might seem difficult to find a pattern in equations (1), (2), (3), and (4), there is indeed one single universal convolution formula that holds for all the Fibonacci, Pell, Padovan, and Tribonacci numbers (and more), so long as we: choose the right initial values for our first sequence, define an appropriate companion sequence, and adjust the limits on the summation. The universal formula will be n−1 X FiLn−i = (n − 1)Fn; (5) i=0 where fFng refers to any Fibonacci-type sequence such as the Fibonacci, Pell, Padovan, Tribonacci, etc., and fLng refers to the companion Lucas-type se- quence; in equations (1), (2), and (3) that would be the Lucas, Pell-Lucas, and 1 Perrin sequences. This universal convolution formula will produce the equations (1), (2), (3), and (4), along with innumerable others. But before we formally state and prove our formula, let us establish some definitions. For a given set C = fc1; : : : ; ckg of complex numbers with ck 6= 0, we define C the associated Fibonacci-type sequence fFn g, henceforth written as just fFng, to be the sequence with values Fn = 0 for n ≤ 0, F1 = 1, and satisfying the recurrence Fn = c1Fn−1 + c2Fn−2 + ··· + ckFn−k for all n ≥ 2. (6) C As for the companion Lucas-type sequence fLng, henceforth written as just fLng, we define it in a slightly different manner: inspired by the Binet formula for the \original" Lucas numbers Ln, we set Ln equal to the sum of the nth powers of the k roots of the characteristic polynomial ∗ k k−1 k−2 h (x) = x − c1x + c2x + ··· + ck (7) corresponding to the recurrence in equation (6). Hence, Ln satisfies the same recurrence as does Fn in equation (6) but without any restrictions on n. For reasons to become clear in a moment, we also need to define another function h(x) to be the reciprocal polynomial for h∗(x), such that h(x) = xkh∗(1=x). In other words, 2 k h(x) = 1 − c1x + c2x + ··· + ckx : (8) An alternate approach to defining Fn and Ln which works well for c1; c2;::: non-negative integers, is to define Fn to be the number of ways to tile a strip of length n − 1 with c1 colors of squares, c2 colors of dominos, and so on, and likewise Ln for tiling a bracelet of length n, with the additional stipulations that F0 = 1 and L0 = k. It is possible to show that this approach gives us the same sequences as the earlier definitions. A third approach would be to define these sequences in terms of their generating functions which we introduce in the proof of Theorem 1, below. We feel that our approach of using initial values to define Fn and a Binet-type formula to define Ln is the most natural of these three options. Let us now consider a few examples and show how they correspond to the sequences that appear in equations (1), (2), (3), and (4). 1. For the set C = fc1; c2g = f1; 1g, our definition of fFng for this set C gives us that F0 = 0, F1 = 1, and Fn = Fn−1 + Fn−2, which are the \original" Fibonacci numbers Fn as seen in equation (1). As for the sequence fLng for this set C, since the two rootsp of the characteristicp polynomial x2−(x+1) from equation (7) are (−1+ 5)=2 and (−1− 5)=2, then we have p !n p !n −1 + 5 −1 − 5 L = + ; n 2 2 2 and we recognize this as the Binet formula for the \original" Lucas num- bers Ln. 2. For C = f2; 1g, our definition tells us that our associated Fibonacci-type numbers for this set C satisfy F0 = 0, F1 = 1, and Fn = 2Fn−1 + Fn−2, producing the sequence 0; 1; 2; 5; 12; 29;::: , and for n ≥ 1 these match the Pell numbers Pn from equation (2) as defined in Koshy [4]. Our Lucas- type numbers Ln in this situation differ slightly from Koshy's Pell-Lucas numbers Qn in equation (2) and in [4]. Our definition gives us p p n n Ln = (1 + 2) + (1 − 2) for n ≥ 0, producing the sequence 2; 2; 6; 14; 34;::: , and when compared to Koshy's definition of Qn we find that our Ln = 2Qn for n ≥ 1. 3. For C = f0; 1; 1g, our Fibonacci-type numbers Fn are 0; 1; 0; 1; 1; 1; 2; 2; 3;::: , while the Padovan numbers An from equation (3) and as defined in [7, 9, 11] have the same recurrence but begin with 1; 1; 1; 2; 2; 3;::: , and so we see that in this case Fn = An−3 for n ≥ 3. The Perrin numbers En from equation (3) and [7, 9, 11] begin with 3; 0; 2; 3; 2; 5; 5; 7;::: and are identical to our Lucas-type numbers Ln in this case. 4. Finally, for C = f1; 1; 1g, our numbers Fn in this case match perfectly with the Tribonacci numbers Tn as defined in Komatsu and Li [3] and as seen in equation (4). The numbers Un = Tn + Tn−2 + 2Tn−3 for n ≥ 3 also match fairly well with our companion Lucas-type numbers Ln in this case. To be precise, we have Un = Ln−1 for n ≥ 3, which can be verified quite easily with some quick calculations with the three roots of x3 − (x2 + x + 1) = 0. This sequence fLng is 3; 1; 3; 7; 11; 21; 39;::: , and is sometimes identified as the Tribonacci-Lucas, Lucas 3-step, or 3-Lucas numbers as seen in [1, 6]. We are now ready to state and prove our main theorem. Theorem 1. For a given set C = fc1; c2; : : : ; ckg with associated Fibonacci-type sequence fFng and companion Lucas-type sequence fLng as defined above, we have n−1 X FiLn−i = (n − 1)Fn: (9) i=0 Proof. Our proof will have three parts. First, we will show that the generating 0 functions for fFng and fLng are f(x) = x=h(x) and g(x) = k − xh (x)=h(x) respectively, with h(x) as defined in equation (8). Then, we will show that the product f(x)g(x) equals (k − 1)f(x) + xf 0(x). And finally, we will use that equality to establish our convolution equation (9). First, the generating functions. To show that the generating function for P1 n fFng is x=h(x), we let f(x) = n=0 Fnx . Since we have defined Fn = 0 for P1 n n ≤ 0, we can extend this sum to −∞ as well, giving us f(x) = n=−∞ Fnx . We now consider f(x)h(x) with h(x) from equation (8). The coefficient of xn 3 in f(x)h(x) is Fn − (c1Fn−1 + c2Fn−2 + ··· + ckFn−k), and for n ≥ 2 this is zero thanks to equation (6). Likewise, for n ≤ 0 this is zero since every Fi = 0 for i ≤ 0. As for n = 1, the expression F1 − (c1F0 + c2F−1 + ··· + ckF−(k−1)) simplifies to 1−(0+0+···+0) = 1. Thus, f(x)h(x) = 1x and so f(x) = x=h(x) as desired. 0 To show that the generating function for fLng is k − xh (x)=h(x), we begin P1 n by defining g(x) = n=0 Lnx . Since L0 = k, then if we let γ(x) = g(x) − k P1 n then γ(x) = n=1 Lnx . We now consider γ(x)h(x), with h(x) from equation n (8). For n ≥ k + 1, the coefficient of x in γ(x)h(x) is Ln − (c1Ln−1 + c2Ln−2 + ··· + ckLn−k) which equals zero since by our discussion earlier, Ln satisfies the same recurrence relation as Fn in equation (6). For 1 ≤ n ≤ k, the situation is more interesting. In this case, the coefficient of xn is Ln − (c1Ln−1 + c2Ln−2 + c3Ln−3 + ··· + cn−1L1); (10) ∗ and since each Ln is defined to be the nth power sum pn of the roots of h (x), then expression (10) becomes pn − (c1pn−1 + c2pn−2 + c3pn−3 + ··· + cn−1p1): (11) ∗ We now recognize that the coefficients c1; c2; : : : ck of h (x) can be written in terms of the elementary symmetric polynomials e1; e2; : : : ek by the simple equa- i+1 tion ci = (−1) ei, and so expression (11) becomes n pn − (e1pn−1 − e2pn−2 + e3pn−3 + ··· + (−1) en−1p1); (12) n+1 and by Newton's identities this is simply (−1) nen which is ncn.
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