Several Generating Functions for Second-Order Recurrence Sequences
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1 47 6 Journal of Integer Sequences, Vol. 009), Article Several Generating Functions for Second-Order Recurrence Sequences István Mező Institute of Mathematics University of Debrecen Hungary imezo@math.lte.hu Abstract Carlitz and Riordan began a study on closed form of generating functions for powers of second-order recurrence sequences. This investigation was completed by Stănică. In this paper we consider eponential and other types of generating functions for such sequences. Moreover, an etensive table of generating functions is provided. Introduction The Fibonacci sequence, which is is sequence A00004 in Sloane s Encylopedia, [] is defined recursively as follows: F n = F n + F n n ), with initial conditions F 0 = 0, F =. The Lucas numbers L n, which comprise Sloane s sequence A0000, are defined by the same manner but with initial conditions L 0 =, L =. In 96, Riordan [9] determined the generating functions for powers of Fibonacci numbers: f ) = Fn n.
2 This question had been suggested by Golomb [4] in 97. Riordan found the recursive solution [/] L + ) )f ) = + ) j A j f j ) j ), ) with initial functions j= f 0 ) =, and f ) =. We mention that in his paper Riordan used the F 0 = F = condition. In the result above, the coefficients A j have a complicated definition and cannot be handled easily. In the same journal and volume, Carlitz [] made the following generalization. Let with initial conditions u n = pu n qu n n ), u 0 =, u = p. He computed the generating functions for the sequences u n. They have the same form as Eq. ). In his recent paper [], Stănică gave the most general and simple answer for the questions above see Theorem ) with an easy proof. Namely, let the so-called second-order recurrence sequence be given by u n = pu n + qu n n ), ) where p,q,u 0 and u are arbitrary numbers such that we eliminate the degenerate case p + 4q = 0. Then let α = p + p + 4q), β = p p + 4q), ) It is nown that u n can be written in the form A = u u 0 β α β, B = u u 0 α α β. 4) u n = Aα n Bβ n Binet formula). Many famous sequences have this shape. A comprehensive table can be found at the end of the paper. To present Stănică s result, we need to introduce the sequence V n given by its Binet formula: V n = α n + β n, V 0 =,V = p. Theorem Stănică). The generating function for the rth power of the sequence u n is u r n n =
3 ) r A ) A B r B r + b) B r α r A r β r ), b) V r r =0 if r is odd, and u r n n = r ) r B ) A B r + A r b) B r α r + A r β r ) b) V r + =0 ) r r A B) + r r ) r, if r is even. In the spirit of this result we present the same formulas for even and odd indices, eponential generating functions for powers, product of such sequences and so on. Non-eponential generating functions The result in this and the following sections yield rich and varied eamples which are collected in separate tables at the end of the paper. First, the generating function for u n is given: Proposition. We have u n n = u 0 + u pu 0 ) p q. For the sae of a more readable presentation, the proof of this statement and all of the others will be collected in a separate section. We remar that Proposition is not new but the proof is easy and typical. Sequences with even and odd indices appear so often that it is worth to construct the general generating function of this type. Theorem. The generating function for the sequence u n is u n n = u 0 + u u 0 p + q)) p + q) + q, while u n+ n = u + u 0 pq u q) p + q) + q.
4 Eample 4. As a consequence, we can state the following identity which we use later. F n n = See the paper of Johnson [6], for eample. +. Generating functions for powers of even and odd indices are interesting. The following theorem contains these results. Theorem. Let u n = pu n + qu n be a sequence with initial values u 0 and u. Then u r n n = ) r E ) E F r F r + q F r ρ r E r σ r ), q V r r =0 if r is odd, and u r n n = r ) r F ) E F r + E r q F r ρ r + E r σ r ) q V r + =0 ) r r E F) + r r ) r, if r is even. For odd indices we have to mae the substitution E G and F H. Here ρ = p + q + p ) p + 4q, σ = p + q p ) p + 4q, E = u u 0 σ ρ σ, F = u u 0 ρ ρ σ, G = u u σ ρ σ, H = u u ρ ρ σ, V n = ρ n + σ n, V 0 =, V = p + q. Remar 6. These constants are calculated for the named sequences: 4
5 Sequence ρ σ E F G H F n + L n + P n Q n + J n 4 φ φ 0 + ) + ) 4 + ) 0 ) ) 4 ) j n 4 The product of the sequences u n and v n has a simple generating function as given in the following proposition. Proposition 7. Let u n and v n be two second-order recurrence sequences given by their Binet formulae: u n = Aα n Bβ n, v n = Cγ n Dδ n, where A, B, C, D, α, β, γ, δ are defined as in Eqs. ) and 4). Then the generating function for u n v n is u n v n n = AC αγ AD αδ BC βγ + BD βδ. We mention that a similar statement can be obtained for the products u n v n, u n v n, u n+ v n, u n+ v n+, etc. Remar 8. As a special case, let u n = F n and v n = L n. Then it is well nown from Binet formula, for eample) that A = B =, α = + C =, D =, γ = + β =, δ =. The quantity + is called the golden ratio or golden mean, or golden section). For further use we apply the standard notation φ for this, and φ for. We remar that φφ = and φ φ = φ φ =. Therefore ) F n L n n = φ φ = φ φ ) φ ) φ ) = +. Comparing the result obtained in Eample 4, this yields the nown identity See Mordell s boo [8, pp. 60 6]. F n = F n L n.
6 Remar 9. The variation u n = F n and v n = P n where P n are the Pell numbers A0009) also have combinatorial sense. See the paper of Sellers [0]. The generating function for F n P n is nown A008, but it can be deduced using the proposition above: F n P n n = Using Theorem, the eponential generating function for F n P n is derived: n F n P n n! = [ ] e φ+) e φ ) e φ+) + e φ ). 4 Remar 0. As the author realized, the sequence J n j n ) appears in the on-line encyclopedia [] but not under this identification J n and j n are called Jacobsthal A0004 and Jacobsthal-Lucas A04 numbers). The sequence A0040 has the generating function as J n j n ). Thus, the definition of A0040 gives the otherwise elementary but not depicted) observation J n j n = 4n. Let us turn the discussion s direction to the determination of generating functions with coefficients un. We do not restrict ourselves to the case of positive q.) To do this, we present n q the notion of polylogarithms which are themselves generating functions, having coefficients. Concretely, n q n Li q ) = n. q Because of the coefficients n q, it is etremely difficult to find closed forms of these sums but the situation changes when we tae negative powers: n q n = Li q ) = ) q+ q i=0 q q i, i where the symbol a b denotes the Eulerian numbers; that is, a b is the number of permutations on the set,...,a in which eactly b elements are greater than the previous element []. After these introductory steps, we state the following. Proposition. For any u n second-order recurrence sequence and for any q Z, In particular, if q = then u n n q n = A Li q α) B Li q β). u n n n = A ln α) + B ln β), ) 6
7 while for q = nu n n = A α) B β). Applications can be found at the end of the paper. We mention that the special case u n = F n and = was investigated by Benjamin et al. [] from a probabilistic point of view. Moreover, we can easily formulate the parallel results for even and odd indices: u n n q n = E Li q ρ) F Li q σ), u n+ n = G Li n q q ρ) H Li q σ). Remar. In their paper on transcendence theory, Adhiari et al. [] noted the beautiful fact that the sum F n n n is transcendental. Possessing the results above, we are able to tae a closer loo at this sum. Let u n = F n and = in Eq. ). Then F n n = ln φ ) + ln φ ) n = ) + )) ln + ln = + ) ln 4 4 = ) φ ln. φ So, this value is a transcendental number. A similar calculation shows that L n n n = ln), which is again a transcendental number. In addition, we present an interesting eample for series whose members denominators and the sum are the same but the numerators are different. Namely, L n n = ln) = n n n. Finding closed form for different arguments of polylogarithms is an intensively investigated and very hard topic. Fortunately, some functional equations gives the chance to find a closed form for the sum of certain series involving Fibonacci and Lucas numbers. In the boo [7, pp. 6 7, 7 9] of Lewin, these are all the nown special values: 7
8 Li ) = π, 6 Li ) = π ), Li = π log ), Li φ) = log φ) π, Li φ) = log φ) + π, 0 Li φ) = log φ) ) π, 0 Li = π φ 4 log ), φ Li ) = ζ), 4 Li ) = 7 π ζ) log) log ), ) ) Li = 4 π ζ) + log φ φ log ). φ Here ζ) = Li ) is the Apéry s constant without nown closed form. With these identities, we deduce the following beautiful sums: Using Proposition, ) n F n φ n n = ) n L n φ n n ) n F n φ n n = ) log φ) π, 0 = log φ) π 60, π logφ) log φ) 0 ζ) ) n L n = π ζ) φ n n 0 logφ) + log φ). ) n F n φ n n = ), F n n φn = Li φφ) Li φφ) = Li ) Li φ )), since φ =. Using the table of polylogarithms above, an elementary calculation shows the φ result. The same approach can be applied to derive the other sums with data from the table with respect to A,B,α,β). We can rewrite these sums in a more curious form, because π = sin. 0) 8
9 That is, Whence, for eample, π ) φ = φ = sin. 0 ) n F n φ n n = ) n F n π ) sin n. n 0 Eponential generating functions The results in the section above can also have eponential versions, which we give net. Since such epressions often cannot be simplified and finding the eponential generating function is only a substitution of constants, we omit the tables. Theorem. The recurrence sequence u n has the eponential generating function n u n n! = Aeα Be β, while for even and odd indices n u n n! n u n+ where E, F, G, H, ρ, σ are defined in Theorem. n! = Ee ρ Fe σ, = Ge ρ He σ, We phrase the eponential version of Stănică s theorem in a wider sense. Theorem 4. We have u r n r n n! = =0 u r n r n n! = =0 u r n r n+ n! = =0 ) r A B) r e α β r, ) r E F) r e ρ σ r, ) r G H) r e ρ σ r. The eponential generating function for product of recurrence sequences is presented in the following Theorem. Under the hypotheses of Proposition 7, we have n u n v n n! = ACeαγ ADe αδ BCe βγ + BDe βδ. Again, the same statement can be obtained for the products u n v n, u n v n, u n+ v n, u n+ v n+, etc. 9
10 4 Proofs Proof of Proposition. Let the generating function be f). Then f) pf) q f) = u 0 + u pu 0 + = u 0 + u pu 0, u n pu n qu n ) n by Eq. ). The result follows. Proof of Theorem. In order to reach our aim, we need the following identity: n= u n = p + q)u n q u n 4. 6) Since we get that u n = pu n + qu n, and u n = pu n + qu n 4, u n = q u n pu n ) = p u n qu n 4 ). If we epress u n and consider the identity u n = p u n qu n ), we will arrive at Eq. 6). Let f e ) be the generating function for u n e abbreviates the word even ). Then = n= q f e ) p + q)f e ) + f e ) = q u n 4 n p + q) u n n + u n n q u n 4 p + q)u n + u n ) n p + q)u 0 + u 0 + u. n= = u 0 + u u 0 p + q)). We get the result. Let f o ) be the generating function for the sequence u n+. pf o ) + qf e ) = pu n+ + qu n ) n = u n+ n Thus = u n n = ) u n n u 0 = f e) u 0 ). f o ) = p ) f e ) q u ) 0. 0
11 If we consider the closed form of f e ) this formula can be transformed into the wanted form. Proof of Theorem. We now see Eq. 6)), that u n = p + q)u n q u n 4. This allows us to construct a second-order recurrence sequence v n from u n with the property v n = u n, namely, Therefore ρ = v n := p + q)v n q v n, v 0 := u 0, v := u. 7) p + q + ) p + q) + 4 q ) = p + q + p ) p + 4q, σ = p + q ) p + q) + 4 q ) = p + q p ) p + 4q, E = v v 0 σ ρ σ = u u 0 σ ρ σ, with respect to the sequence v n. That is, v n = Eρ n Fσ n. F = v v 0 ρ ρ σ = u u 0 ρ ρ σ If we apply Stănică s theorem for v n = u n, we get the first statement. Secondly, we find the corresponding identity of Eq. 6). We epress u n from these: u n = pu n + qu n, and u n = pu n + qu n. u n = p u n qu n ) = q u n pu n ), whence On the other hand, u n = q p u n qu n ) + pu n. u n = p u n+ qu n ). Putting together the last two equalities we get the wanted formula: u n+ = p + q)u n q u n. 8) Again, we are able to construct the sequence w n for which w n = u n+.
12 We are in the same situation as before. The only thing we should care about is that w 0 = u, w = u. Thus Proof of Proposition 7. If u n and v n have the form as in the proposition, then we see that = AC u n v n = ACαγ) n ADαδ) n BCβγ) n + BDβδ) n. αγ) n AD u n v n n αδ) n BC βγ) n + BD βδ) n. The result follows. In addition, we mention that there are too many parameters, so it is not worth to loo for an epression with parameters u 0,u,v 0,v,p,q,r,s directly. However, the remains can be completed easily, as the author calculated for the standard sequences. Proof of Proposition. It is straightforward from Binet formula and the definition of polylogarithms. Proof of Theorem. This proof is again straightforward, u n n n! = A α) n n! β) n B n! = Ae α Be β. Finally, we choose v n and w n as in the proof of Theorem, and follow the usual argument. Proofs of Theorems 4 and. The binomial theorem, the same approach as described in the proof of Theorem and the Binet formula immediately gives the results: u r n n n! = = = = Aα n Bβ n ) rn n! r r )A α ) n B) r β r ) nn n! =0 r ) r A B) r α ) n β r ) nn n! r ) r A B) r e α β r. =0 =0 The rest can be proven by the same approach.
13 Tables Standard parameters for the named sequences Name Notation u 0 u p q First few values Fibonacci F n 0 0,,,,,, 8,, Lucas L n,,, 4, 7,, 8, 9, 47 Pell P n 0 0,,,,,9,70,69,408 Pell-Lucas Q n,,6,4,4,8,98,478 Jacobsthal J n 0 0,,,,,,, 4, 8 Jacobsthal-Lucas j n,,,7,7,,6,7,7 Sequence A B α β F n + L n + P n Q n + J n j n Ordinary generating functions Coefficient of n F n L n P n Q n J n j n Generating function
14 Generating functions of even and odd indices n Generating function n Generating function F n L n P n Q n J n j n F + n+ + + L + n+ + P 6+ n Q 6+ n+ 6+ J +4 n j +4 n+ +4 Generating functions for products of sequences Coefficient of n Generating function F n L n + F n P n F n Q n F n J n F n j n L n P n L n Q n L n J n L n j n P n Q n 6+ P n J n P n j n Q n J n Q n j n J n j n +4 4
15 Generating functions for squares n Gen. function n Gen. function n Gen. function Fn L n Pn Q n Jn jn Fn L n Pn Q n Jn jn F n L n Pn Q n J n j n Generating functions for sequences n u )n ) n Gen. function n Gen. function + nf n nf n nl 4 n nl n np n np n nq 6 4 n nq n nj 4 n nj n nj 6 n nj n Generating functions for sequences n u n+ ) Coefficient of n Generating function nf n nl n np n nq n nj n nj n
16 References [] S. D. Adhiari, N. Saradha, T. N. Shorey and R. Tijdeman, Transcendental infinite sums, Indag. Math. J. ) 00), 4. [] A. T. Benjamin, J. D. Neer, D. E. Otero, J. A. Sellers, A probabilistic view of certain weighted Fibonacci sums, Fib. Quart. 4 4) 00), [] L. Carlitz, Generating functions for powers of certain sequences of numbers, Due Math. J. 9 96), 7. [4] S. Golomb, Problem 470, Amer. Math. Monthly 64) 97), p. 49. [] R. L. Graham, D. E. Knuth and O. Patashni, Concrete Mathematics, Addison Wesley, 99. [6] R. C. Johnson, Matri method for Fibonacci and related sequences, notes for undegraduates, 006. Available online at [7] L. Lewin, Dilogarithms and Associated Functions, Macdonald, London, 98. [8] L. J. Mordell, Diophantine Equations, Academic Press, London and New Yor, 969. [9] J. Riordan, Generating functions for powers of Fibonacci numbers, Due Math. J. 9 96),. [0] J. A. Sellers, Domino tilings and products of Fibonacci and Pell Numbers, J. Integer Seq. ), 00), Article 0... [] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences. Published electronically at [] P. Stănică, Generating functions, weighted and non-weighted sums for powers of secondorder recurrence sequences, Fib. Quart. 44) 00),. 000 Mathematics Subject Classification: Primary B9. Keywords: Recurrence sequences, Fibonacci sequence, Lucas sequence, Pell sequence, Pell- Lucas sequence, Jacobsthal sequence, Jacobsthal-Lucas sequence, generating function, eponential generating function. Concerned with sequences A0000, A00004, A0009, A0004, A008, A0040, and A04.) Received July 7 007; revised version received April Published in Journal of Integer Sequences, April Return to Journal of Integer Sequences home page. 6
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