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 {Fn} refers to any Fibonacci-type sequence such as the Fibonacci, Pell, Padovan, Tribonacci, etc., and {Ln} refers to the companion Lucas-type sequence; 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 = {c1, . . . , ck} of complex numbers with ck 6= 0, we deﬁne C the associated Fibonacci-type sequence {Fn }, henceforth written as just {Fn}, 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 {Ln}, henceforth written as just {Ln}, 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 = {c1, c2} = {1, 1}, our deﬁnition of {Fn} 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 {Ln} for this set C, since the two roots√ of the characteristic√ polynomial x2−(x+1) from equation (7) are (−1+ 5)/2 and (−1− 5)/2, then we have √ !n √ !n −1 + 5 −1 − 5 L = + , n 2 2

2 and we recognize this as the Binet formula for the “original” Lucas numbers Ln. 2. For C = {2, 1}, 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 √ √ 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 = {0, 1, 1}, 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 = {1, 1, 1}, 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 {Ln} 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 = {c1, c2, . . . , ck} with associated Fibonacci-type sequence {Fn} and companion Lucas-type sequence {Ln} 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 {Fn} and {Ln} 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 P∞ n {Fn} is x/h(x), we let f(x) = n=0 Fnx . Since we have defined Fn = 0 for P∞ 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 {Ln} is k − xh (x)/h(x), we begin P∞ n by defining g(x) = n=0 Lnx . Since L0 = k, then if we let γ(x) = g(x) − k P∞ 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. Hence, since n the coefficient of x in γ(x)h(x) is ncn for 1 ≤ n ≤ k and is zero fo n ≥ k + 1, then 2 3 k γ(x)h(x) = c1x + 2c2x + 3c3x + ··· + kckx , (13) and since the right-hand side of equation (13) is simply −xh0(x), then γ(x) = −xh0(x)/h(x) and so we have g(x) = k + γ(x) = k − xh0(x)/h(x), as desired. Next, the product f(x)g(x). To assist us in this part, let us first recognize that ln f(x) = ln x − ln h(x). Taking the derivative of both sides, we find that f 0(x) 1 h0(x) = − . (14) f(x) x h(x) If we multiply both sides by x, we obtain xf 0(x) xh0(x) = 1 − , (15) f(x) h(x) and if we add k − 1 to both sides and simplify, we have (k − 1)f(x) + xf 0(x) xh0(x) = k − , (16) f(x) h(x)

4 and we recognize that the right-hand side of equation (16) is g(x), so after multiplying both sides by f(x) we obtain our desired equation,

(k − 1)f(x) + xf 0(x) = f(x)g(x). (17)

P n P n Finally, the convolution. We recall that f(x) = Fnx and g(x) = Lnx , n and so the coeﬃcient of x in the left-hand side of equation (17) is (k − 1)Fn + n nFn. Likewise, the coeﬃcient of x in the right-hand side of equation (17) is Pn i=0 FiLn−i. Comparing the two, we have

n X (k − 1)Fn + nFn = FiLn−i. (18) i=0 If we now subtract the last term in that summation from both sides, we have

n−1 X (k − 1)Fn + nFn − FnL0 = FiLn−i, (19) i=0 and since L0 = k then the left-hand side of (19) simpliﬁes to (n − 1)Fn, thus giving us our desired equation (9). We can now use Theorem 1 to establish equations (1), (2), (3), and (4) from the beginning of this paper. For equation (1), we know from our discussion earlier that when {c1, c2} = {1, 1} then Fn equals the “original” Fibonacci number Fn and likewise Ln is the “original” Lucas number Ln. Thus, equation (9) becomes n−1 X Fi Ln−i = (n − 1)Fn, i=0 and if we add 2Fn to both sides we obtain equation (1). Likewise, for equation (2) we recall that when C = {c1, c2} = {1, 2} then Fn = Pn and Ln = 2Qn for Pn and Qn the Pell and Pell-Lucas numbers. Hence, equation (9) in this case becomes n−1 X Pi 2Qn−i = (n − 1)Pn, i=0 and if we divide both sides by 2 we obtain equation (2). As for equation (3), our earlier discussion established that for C = {0, 1, 1} then Fn = An−3 (for n ≥ 3) and Ln = En, with An and En the Padovan and Perrin numbers. To derive equation (3), we begin with equation (9) but with n + 3 instead of n and we separate out the ﬁrst three terms to obtain

n+2 X F0Ln+3 + F1Ln+2 + F2Ln+1 + FiL(n+3)−i = (n + 2)Fn+3. i=3

5 If we now substitute F0 = 0, F1 = 1, F2 = 0, Ln = En, and Fn = An−3 (this last one for n ≥ 3) then the above equation becomes

n+2 X En+2 + Ai−3En+3−i = (n + 2)An, i=3 and if we re-index the summation then this becomes

n−1 X En+2 + AiEn−i = (n + 2)An. i=0

It remains only to move the En+2 to the other side, and add AnE0 = 3An to both sides, to obtain equation (3). Finally, for the Tribonacci convolution in equation (4), recall from our earlier discussion that with C = {1, 1, 1} then the Tribonacci numbers Tn do indeed equal our Fn but the numbers Un in equation (4) equal Ln−1 and only for n ≥ 3. Thus, to establish (4) we begin with equation (9) but with n − 1 instead of n and we separate out the last term to obtain

n−3 X FiL(n−1)−i + Fn−2L1 = (n − 2)Fn−1. i=0

We can now use the identities L1 = 1, Fn = Tn, and for n ≥ 3 that Un = Ln−1 to quickly obtain equation (4). We point out that many other authors have studied the convolution of Fibonacci-type and Lucas-type numbers. Zeitlin [12, Formula (5.5)] gave a version of Theorem 1 for second-order recurrences; his formula is

n X Zn+1−iVi+1 = (n + 1)Zn+2, (20) i=0 where his Zn is the Fibonacci-type sequence that satisfies the recurrence formula Zn = aZn−1+bZn−2, and likewise his Vn is the Lucas-type sequence that satisfies the same recurrence. In our notation, we would say that Zeitlin covered the cases where C = {a, b}. Robbins [8, Theorem 5] improved on Zeitlin’s result with an identity that holds for weighted convolutions of Zeitlin’s Zn and Vn sequences. Szakács[10, Theorem 5] also covered second-order recurrences and gave specific examples for the Fibonacci, Pell, and Jacobsthal sequences. Dresden and Wang [1, Theorem 5] gave a induction proof of a version of Theorem 1 that only applied to the so-called k-bonacci and k-Lucas numbers corresponding to the specific sets C = {1, 1,..., 1}. Finally, we note that Koshy and Griffiths [5, equation (2.2)] discovered this delightful convolution formula that connects the seemingly-unrelated Jacobsthal numbers Jn and Fibonacci numbers Fn,

n X JiFn−i = Jn+1 − Fn+1, (21) i=0

6 and Frontczak [2, Theorem 2.1] found an equally delightful formula that links the Tribonacci numbers Tn with the Fibonacci numbers Fn,

n X TiFn−i = Tn+2 − Fn+2. (22) i=0

We can’t help but notice the similarity between equations (21) and (22), and we can’t help but ask if there are other universal convolution formulas waiting to be discovered.

References

[1] Greg Dresden and Yichen Wang. Sums and convolutions of k-bonacci and k-Lucas numbers. Integers, 21:Paper No. A56, 16, 2021. [2] Robert Frontczak. Some Fibonacci-Lucas-Tribonacci-Lucas identities. Fi- bonacci Quart., 56(3):263–274, 2018. [3] Takao Komatsu and Rusen Li. Convolution identities for tribonacci numbers with symmetric formulae. Math. Rep. (Bucur.), 21(71)(1):27–47, 2019. [4] Thomas Koshy. Pell and Pell-Lucas numbers with applications. Springer, New York, 2014. [5] Thomas Koshy and Martin Griﬃths. Some Gibonacci convolutions with dividends. Fibonacci Quart., 56(3):237–245, 2018. [6] Tony D. Noe and Jonathan Vos Post. Primes in Fibonacci n-step and Lucas n-step sequences. J. Integer Seq., 8(4):Article 05.4.4, 12, 2005.

[7] Salah Eddine Rihane, ChèfiathAwero Adegbindin, and Alain Togbé. Fer- mat Padovan and Perrin numbers. J. Integer Seq., 23(6):Art. 20.6.2, 11, 2020. [8] Neville Robbins. Some convolution-type and combinatorial identities per- taining to binary linear recurrences. Fibonacci Quart., 29(3):249–255, 1991. [9] Ian Stewart. Tales of a neglected number. Sci. Amer., 274(6):102–103, 1996. [10] TamásSzakács.Convolution of second order linear recursive sequences II. Commun. Math., 25(2):137–148, 2017.

[11] Steven J. Tedford. Combinatorial identities for the Padovan numbers. Fi- bonacci Quart., 57(4):291–298, 2019. [12] David Zeitlin. On convoluted numbers and sums. Amer. Math. Monthly, 74:235–246, 1967.