LINEAR RECURRENCE SEQUENCES VIA BELL POLYNOMIALS

arXiv:1405.7727v1 [math.CO] 29 May 2014

LINEAR RECURRENCE SEQUENCES VIA BELL POLYNOMIALS DANIEL BIRMAJER, JUAN B. GIL, AND MICHAEL D. WEINER

Abstract. For arbitrary homogeneous linear recurrence sequences with fixed coefficients we give a representation in terms of partial Bell polynomials. Using known properties of these polynomials several combinatorial identities are obtained, including formulas for multifold convolutions of certain fundamental sequences.

1. Introduction A linear recurrence sequence (an ) of elements in a commutative ring R is a sequence given by a homogeneous linear recurrence relation an = c1 an−1 + c2 an−2 + · · · + cd an−d for n ≥ d,

(1.1)

with fixed coefficients c1 , . . . , cd ∈ R, together with initial values a0 , a1 , . . . , ad−1 ∈ R. The aim of this note is to derive the representation an =

d−1 X k=0

λk

n−k X j=0

j! Bn−k,j (1!c1 , 2!c2 , . . . ) for n ≥ 1, (n − k)!

(1.2)

where Bn,k = Bn,k (x1 , x2 , . . . ) denotes the (n, k)-th partial Bell polynomial, and the λk ’s are constants (given explicitly in Theorem 2.1) depending on the initial values of the sequence. In order to motivate this formula, recall that the generating function of a linear recurrence sequence given by (1.1) is a rational function f (x) = p(x) q(x) with q(x) = 1 − c1 x − c2 x2 − · · · − cd xd , and a polynomial p(x) of degree at most d − 1 that depends on the initial values. In other words, the function f (x) is a linear combination of the rational functions 1 x x2 xd−1 , , , ..., , q(x) q(x) q(x) q(x) and their respective associated sequences form a basis for the space of linear recurrence sequences with coefficients c1 , . . . , cd . Note that the sequence with generating function 1

2

DANIEL BIRMAJER, JUAN B. GIL, AND MICHAEL D. WEINER

1/q(x) corresponds to the sequence satisfying the relation (1.1) with initial values a0 a1 a2 a3

ad−1

=1 = c1 = c1 a1 + c2 = c1 a2 + c2 a1 + c3 .. . = c1 ad−2 + c2 ad−3 + · · · + cd−2 a1 + cd−1 .

Motivated by these facts, and in an attempt to investigate linear recurrence sequences all at once (regardless of length), we replace q(x) by the power series Q(x) = 1 −

∞ X

cn xn

n=1

and consider its reciprocal Y (x) = P 1/Q(x). Using Fa`a di Bruno’s formula, it follows that n Y (x) can be written as Y (x) = 1 + ∞ n=1 yn x with yn =

n X k! k=0

n!

Bn,k (1!c1 , 2!c2 , 3!c3 , . . . )

(1.3)

(cf. Theorem B in [4, Section 3.5]). Moreover, since Q(x) · Y (x) = 1, one can easily check the recurrence relation n X ck yn−k for n ≥ 1. (1.4) yn = k=1

Note that, if there are only finitely many coefficients c1 , . . . , cd , then (by construction) the sequence (yn ) is precisely the linear recurrence sequence with generating function 1/q(x). In Section 2 we establish our main result Theorem 2.1 and discuss a variety of illustrating examples, obtaining combinatorial identities by means of basic properties of the partial Bell polynomials. In particular, our approach provides an alternative way to derive the GirardWaring formulas for the power sum symmetric functions. As an application, in Section 3 we give a general convolution formula (Theorem 3.3) for sequences of the form (1.3) that relies on a convolution formula for partial Bell polynomials proved by the authors in [2]. As an immediate consequence of our result, we obtain a (possibly new) multifold convolution identity (3.7) for the Tribonacci numbers. 2. Linear Recurrence Sequences In this section, we will use the sequence (yn ) introduced in (1.3) to write arbitrary linear recurrence sequences in terms of partial Bell polynomials. Theorem 2.1. Let (an ) be a linear recurrence sequence satisfying an = c1 an−1 + c2 an−2 + · · · + cd an−d for n ≥ d ≥ 1,

LINEAR RECURRENCE SEQUENCES VIA BELL POLYNOMIALS

3

with initial values a0 , a1 , . . . , ad−1 . Let (yn ) be the sequence defined by yn =

n X k! k=0

n!

Bn,k (1!c1 , 2!c2 , . . . ) for n ≥ 0.

If we define λ0 , λ1 , . . . , λd−1 by λ0 = a0 and λn = an −

n X

cj an−j for n = 1, . . . , d − 1,

j=1

then an = λ0 yn + λ1 yn−1 + · · · + λd−1 yn−d+1 , hence an =

d−1 X k=0

λk

n−k X j=0

j! Bn−k,j (1!c1 , 2!c2 , . . . ) for n ≥ 1. (n − k)!

Proof. As discussed in the previous section, the sequence (yn ) satisfies the same recurrence relation as (an ), and if S denotes the right-shift operator S(a1 , a2 , . . . ) = (0, a1 , a2 , . . . ), then the sequences (yn ), S(yn ), S 2 (yn ), . . . , S d−1 (yn ) form a basis for the space of all linear recurrence sequences with coefficients c1 , . . . , cd . Thus there are constants λ0 , λ1 , . . . , λd−1 such that an = λ0 yn + λ1 yn−1 + · · · + λd−1 yn−d+1 with the convention that yk = 0 if k < 0. In order to find the λk ’s, we just need to look at the initial values and solve the equation      λ0 a0 1 0 ··· ··· ··· 0  y1 1 0 · · · · · · 0  λ1   a1       y1 1 0 · · · 0  λ2   a2   y2      . ..   ..  =  ..  .. .. .  . . .  .   .  . .       . . .  . .. ..  .. . ..   ..   ..  . yd−1 yd−2 · · · · · · y1 1 λd−1 ad−1 Because of (1.4), the inverse of the above d × d matrix is  1 0 ··· ··· 1 0 ···  −c1  . .. ..  . . .  .  . . ..  .. .. . −cd−1 −cd−2 · · · −c1

 0 0 ..   . , ..  . 1

so the claimed formula follows by applying this matrix to the vector (a0 , . . . , ad−1 ).

In order to illustrate this result, let us consider a few examples. Example 2.2. (k-Fibonacci) For any positive real number k, consider the linear recurrence sequence defined by Fk,n

Fk,0 = 0, Fk,1 = 1, = kFk,n−1 + Fk,n−2 for n ≥ 2.

4


Then c1 = k, c2 = 1, so λ0 = 0, λ1 = 1, and for n ≥ 1 we get Fk,n = yn−1 =

n−1 X i=0

n−1 X i i! Bn−1,i (k, 2, 0, . . . ) = k2i−n+1 . (n − 1)! n−1−i i=0

With the change of variables j = n − 1 − i, this gives the known formula Fk,n =

⌊ n−1 ⌋ 2

X j=0

n − 1 − j n−1−2j k for n ≥ 1. j

(2.3)

Example 2.4. (Lucas) Let p, q ∈ Z, and consider the recurrence sequence U0 = 0, U1 = 1, Un = p Un−1 − q Un−2 for n ≥ 2. Then c1 = p, c2 = −q, so λ0 = 0, λ1 = 1, and for n ≥ 1 we get Un = yn−1 =

n−1 X j=0

=

j! Bn−1,j (p, −2q, 0, . . . ) (n − 1)!

n−1 X j=0

j p2j−n+1 (−q)n−1−j . n−1−j

Again, with the change of variables k = n − 1 − j, we obtain the formula ⌊ n−1 ⌋ 2

Un =

X

k

(−1)

k=0

n − 1 − k n−2k−1 k p q for n ≥ 1. k

(2.5)

Similarly, we consider the related recurrence sequence V0 = 2, V1 = p, Vn = p Vn−1 − q Vn−2 for n ≥ 2. In this case, λ0 = 2, λ1 = −p, and for n ≥ 1 we get Vn = 2yn − pyn−1 n n−1 X X j j 2j−n n−j =2 p (−q) −p p2j−n+1 (−q)n−1−j n−j n−1−j j=0 j=0 n n X X j − 1 j 2j−n n−j =2 p (−q) − p2j−n (−q)n−j , n−j n−j j=0

j=1

which after grouping, and with the substitution k = n − j, gives n

⌊2⌋ X (−1)k Vn = k=0

n n − k n−2k k p q for n ≥ 1. n−k k

(2.6)


5

Example 2.7. (Padovan) Consider the recurrence sequence P0 = 0, P1 = P2 = 1, Pn = Pn−2 + Pn−3 for n ≥ 3. Then c1 = 0, c2 = c3 = 1, so λ0 = 0, λ1 = λ2 = 1, and for n ≥ 2 we get Pn = yn−1 + yn−2 =

n−1 X j=0

=

n−2

X j! j! Bn−1,j (0, 2, 6, 0, . . . ) + Bn−2,j (0, 2, 6, 0, . . . ) (n − 1)! (n − 2)!

n−1 X j=0

j=0

j n − 1 − 2j

+

n−2 X j=0

j , n − 2 − 2j

which gives the identity X

Pn =

⌈ n−2 ⌉≤j≤⌊ n−1 ⌋ 3 2

j+1 n − 1 − 2j

for n ≥ 1.

(2.8)

Example 2.9. (Chebyshev) We now consider the Chebyshev polynomials defined by T0 (x) = 1, T1 (x) = x, Tn (x) = 2xTn−1 (x) − Tn−2 (x) for n ≥ 2. Then c1 = 2x, c2 = −1, so λ0 = 1, λ1 = −x, and for n ≥ 1 we get Tn (x) = yn − xyn−1 =

n−1 n X X j! j! Bn,j (2x, −2, 0, . . . ) − x Bn−1,j (2x, −2, 0, . . . ) n! (n − 1)! j=0

j=0

n X

n−1 X j j 2j−n n−j = (2x) (−1) −x (2x)2j−n+1 (−1)n−1−j n−j n−1−j j=0 j=0 n n X j 1X j−1 2j−n n−j (2x)2j−n (−1)n−j , = (2x) (−1) − 2 n−j n−j

j=1

j=0

which can be written as ⌊n⌋

Tn (x) =

2 X

k=0

n n−k (−1) (2x)n−2k for n ≥ 1. 2(n − k) k k

(2.10)

Similarly, for the Chebyshev polynomials of second kind Un (x), defined by the same recurrence relation as for Tn (x), but with initial values U0 (x) = 1 and U1 (x) = 2x, we get n

⌊2⌋ X n−k (−1)k (2x)n−2k for n ≥ 1. Un (x) = k k=0

(2.11)

6


Power Sums. We finish this section by considering the power sum symmetric functions sn = xn1 + · · · + xnd with d variables. The sequence (sn ) satisfies the relations (Newton’s identities): s0 s1 s2 s3

=d = e1 = e1 s1 − 2e2 = e1 s2 − e2 s1 + 3e3 .. .

sn = e1 sn−1 − e2 sn−2 + · · · + (−1)d−1 ed sn−d for n ≥ d, where e1 , . . . , ed are the elementary symmetric functions in x1 , . . . , xd . In other words, (sn ) is a linear recurrence sequence of length d with initial values s0 , s1 , . . . , sd−1 . Thus, as a consequence of Theorem 2.1, each power sum sn can be expressed in terms of partial Bell polynomials in e1 , . . . , ed . This representation is an efficient way to organize and prove the Girard-Waring formulas (see e.g. [5]). Proposition 2.12. Let sn = xn1 + · · · + xnd and let e1 , . . . , ed be the elementary symmetric functions in the variables x1 , . . . , xd . Then, for n ≥ 1, we have sn =

n X (k − 1)! Bn,k (1!e1 , 2!e2 , . . . , d!ed , 0, . . . ). (−1)n+k (n − 1)! k=1

Our proof relies on the following basic recursive formula. Lemma 2.13. For any sequence x = (x1 , x2 , . . . ), we have n−k+1 X n j xj Bn−j,k−1 (x). nBn,k (x) = j j=1

Proof. This is a consequence of the known identity n−k n−k+1 X n − 1 X n − 1 Bn,k = xj+1 Bn−1−j,k−1 = xj Bn−j,k−1, j j−1 j=0

j=1

see equation (11.11) on p. 415 in [3]. Therefore, n−k+1 n−k+1 X n X n − 1 j n xj Bn−j,k−1 xj Bn−j,k−1 = nBn,k = j j−1 j=1

j=1

as claimed.

Proof of Proposition 2.12. Let sn = xn1 + · · · + xnd . As mentioned before, (sn ) is a linear recurrence sequence with coefficients cj = (−1)j−1 ej for j = 1, . . . , d,


7

and initial values s0 = d,

sk =

k−1 X

cj sk−j + kck for k = 1, . . . , d − 1.

j=1

P According to Theorem 2.1, we can write sn = d−1 j=0 λj yn−j with (yn ) defined as in (1.3) and the λj ’s given by λ0 = s0 = d and X X j j−1 j X ci sj−i = (j − d)cj ci sj−i + jcj − ci sj−i = λj = sj − i=1

i=1

i=1

for j = 1, . . . , d − 1. Also, recall that (yn ) is designed to satisfy the same recurrence relation P as (sn ) for n ≥ d. Therefore, yn = dj=1 cj yn−j and we can rewrite sn as follows sn =

d−1 X

λj yn−j

j=0

X X d d−1 cj yn−j + =d (j − d)cj yn−j j=1

j=1

= dcd yn−d +

d−1 X

jcj yn−j =

=

jcj

j=1 k=0

λj = (j − d)cj

jcj yn−j

j=1

j=1

n−j d X X

d X

k! Bn−j,k (1!c1 , 2!c2 , . . . , d!cd , 0, . . . ). (n − j)!

Since cj = 0 for j > d, we can write the sum over j up to n to facilitate a change of summation. Then n−j n X X k! jcj Bn−j,k (1!c1 , 2!c2 , . . . ) sn = (n − j)! j=1 k=0

=

n−1 X n−k X

=

n−1 X

k=0 j=1

k=0

jcj

k! Bn−j,k (1!c1 , 2!c2 , . . . ) (n − j)!

n−k k! X n j (j!cj )Bn−j,k (1!c1 , 2!c2 , . . . ) j n! j=1

which by Lemma 2.13 yields sn =

n−1 X k=0

n

X (k − 1)! k! nBn,k+1 (1!c1 , 2!c2 , . . . ) = Bn,k (1!c1 , 2!c2 , . . . ). n! (n − 1)! k=1

The claimed identity for the power sum sn follows by replacing back cj = (−1)j−1 ej and using the homogeneity properties of the polynomial Bn,k .

8


3. Convolutions Motivated by Theorem 2.1, in this section we will consider convolutions of polynomial sequences of the form n X k! Yn (x1 , x2 , . . . ) = Bn,k (x1 , . . . , xn−k+1 ). n! k=0

To this end, let us first recall a convolution formula for Bell polynomials proved by the authors in [2, Section 3]. Lemma 3.1. For any α, τ ∈ R, we have the identity k X n X

α k−ℓ

ℓ=0 m=ℓ

τ −α ℓ k ℓ

n m

Bm,ℓ Bn−m,k−ℓ

τ = Bn,k . k

With this in mind, here is our first convolution formula: Lemma 3.2. The sequence (Yn ) satisfies

n X

Ym Yn−m =

m=0

n X (k + 1)! k=0

n!

Bn,k .

Proof. By definition, n X

m=0

Ym Yn−m

n−m n X m X X j! ℓ! Bm,ℓ Bn−m,j = m! (n − m)! m=0 j=0

ℓ=0

n X

X n X k

(k − ℓ)! ℓ! Bm,ℓ Bn−m,k−ℓ = m! (n − m)! m=0 k=0 ℓ=0 X n k X n k−ℓ (−1)ℓ n X (−1) k! k m (−1) Bm,ℓ Bn−m,k−ℓ , = k n! ℓ ℓ=0 m=ℓ k=0 X −1 −1 n k X n n X k! k−ℓ ℓ m k B B (−1) = m,ℓ n−m,k−ℓ , k n! ℓ ℓ=0 m=ℓ

k=0

which by Lemma 3.1 with α = −1, τ = −2, gives the claimed identity. Theorem 3.3. For r ≥ 2, we have X

Ym1 · · · Ymr

m1 +···+mr =n

n X k! k + r − 1 Bn,k . = n! k k=0


9

Proof. We proceed by induction in r. The case r = 2 is discussed in the previous lemma. Assume the formula holds for products of length less than r > 2. X

Ym1 · · · Ymr =

m1 +···+mr =n

n X

m=0

=

n X

Ym

n−m X j=0

n X m X

m=0

Ym1 · · · Ymr−1

m1 +···+mr−1 =n−m

m=0

=

X

Ym

ℓ=0

j! j+r−2 Bn−m,j (n − m)! j

ℓ! Bm,ℓ m!

n−m X j=0

(j + r − 2)! Bn−m,j (r − 2)!(n − m)!

n X n X k X (k − ℓ + r − 2)! ℓ! Bm,ℓ Bn−m,k−ℓ m! (r − 2)!(n − m)! m=0 k=0 ℓ=0 k n k−ℓ+r−2 n n X k! X X k−ℓ m Bm,ℓ Bn−m,k−ℓ = k n! ℓ k=0 ℓ=0 m=ℓ k n n 1−r n X k! X X (−1)k−ℓ k−ℓ m Bm,ℓ Bn−m,k−ℓ = k n! ℓ ℓ=0 m=ℓ k=0 X 1−r −1 n n n k X X k! k−ℓ ℓ m k (−1) Bm,ℓ Bn−m,k−ℓ , = k n! ℓ

=

ℓ=0 m=ℓ

k=0

which by Lemma 3.1 with α = 1 − r, τ = −r, gives n n X X k! k! k + r − 1 k −r = (−1) Bn,k = Bn,k . n! n! k k k=0

k=0

This finishes the proof.

For illustration purposes, let us consider a wide but simple class of linear recurrence sequences of length 2 that are generated by a single basis element. Example 3.4. For arbitrary α and coefficients c1 and c2 , let (an ) be defined by a0 = 0, a1 = α, an = c1 an−1 + c2 an−2 for n ≥ 2. In the terminology of Theorem 2.1, we then have λ0 = 0, λ1 = α, and for n ≥ 1 we get an = αyn−1 = αYn−1 (c1 , 2c2 , 0, . . . ). By Theorem 3.3, we then get X X am1 +1 · · · amr +1 = αr Ym1 · · · Ymr m1 +···+mr =n

=

m1 +···+mr =n n X r k! k +

α

k=0

n!

r−1 Bn,k (c1 , 2c2 , 0, . . . ). k

10


Now, using Bn,k (c1 , 2c2 , 0, . . . ) = we arrive at

2k−n n−k n! k c2 k! n−k c1

X

⌊2⌋ X

and the change of variable j = n − k,

n

am1 +1 · · · amr +1 =

m1 +···+mr =n

αr

j=0

n − j + r − 1 n − j n−2j j c1 c2 . n−j j

(3.5)

Example 3.6. (Tribonacci, A000073 in [6]) Let (tn ) be the sequence defined by t0 = t1 = 0, t2 = 1, an = an−1 + an−2 + an−3 for n ≥ 3. According to Theorem 2.1, we have tn = yn−2 = Yn−2 (1!, 2!, 3!, 0, . . . ), so tn =

n−2 X k=0

n! k!

and since Bn,k (1!, 2!, 3!, 0, . . . ) = n−2 k XX

tn =

k=0 ℓ=0

k k−ℓ

k! Bn−2,k (1!, 2!, 3!, 0, . . . ), (n − 2)! Pk

k ℓ=0 k−ℓ

k−ℓ n+ℓ−2k

k−ℓ n − 2 + ℓ − 2k

=

, we get

n−2 k XX k=0 ℓ=0

k ℓ

ℓ . n−2−k−ℓ

With the change of variable ℓ = j − k, and changing the order of summation, we arrive at tn =

j n−2 XX j=0 k=0

k j−k

j−k . n−2−j

Moreover, we have the convolution formula X

tm1 +2 · · · tmr +2

m1 +···+mr =n

k n X X k+r−1 k k−ℓ . = k k − ℓ n + ℓ − 2k k=0 ℓ=0

The special case r = 2 is listed as sequence A073778 in [6]. (d)

In general, for higher d-step Fibonacci numbers Fn (d)

F0

(d)

satisfying

(d)

= · · · = Fd−2 = 0, Fd−1 = 1,

Fn(d) =

d X

(d)

Fn−j for n ≥ d,

j=1

(d)

we have Fn

m1

= yn−d+1 = Yn−d+1 (1!, 2!, . . . , d!, 0, . . . ), and so n X X k! k + r − 1 (d) (d) Bn,k (1!, 2!, . . . , d!, 0, . . . ). Fm1 +d−1 · · · Fmr +d−1 = n! k +···+m =n r

k=0

(3.7)


11

References [1] E.T. Bell, Exponential polynomials, Ann. of Math. 35 (1934), pp. 258–277. [2] D. Birmajer, J. Gil, and M. Weiner, Some convolution identities and an inverse relation involving partial Bell polynomials, Electron. J. Combin. 19 (2012), no. 4, Paper 34, 14 pp. [3] C.A. Charalambides, Enumerative Combinatorics, Chapman and Hall/CRC, Boca Raton, 2002. [4] L. Comtet, Advanced Combinatorics: The Art of Finite and Infinite Expansions, D. Reidel Publishing Co., Dordrecht, 1974. [5] H. Gould, The Girard-Waring power sum formulas for symmetric functions and Fibonacci sequences, Fibonacci Quart. 37 (1999), no. 2, 135–140. [6] N.J.A. Sloane, The On-Line Encyclopedia of Integer Sequences, https://oeis.org/. Department of Mathematics, Nazareth College, 4245 East Ave., Rochester, NY 14618 Penn State Altoona, 3000 Ivyside Park, Altoona, PA 16601