An Euler-Like Product with Fibonacci Exponents

The pentagonal theorem for partitions is a consequence of the expansion of Euler’s famous product ( 1 − y )( 1 − y 2 )( 1 − y 3 )( 1 − y 4 )( 1 − y 5 ) · · · We investigate the nature of the coefﬁcients of the series expansion of ( 1 − y )( 1 − y 2 )( 1 − y 3 )( 1 − y 5 )( 1 − y 8 ) · · · , in which the sequence of exponents is the Fibonacci numbers. As a part of the study of the combinatorial properties of the development of this product, we show that the series expansion coefﬁcients are from {− 1 , 0 , 1 } , and their behavior is determined by a monoid of twenty-ﬁve 2 × 2 matrices.


Introduction
Once we interpret the famous product of Euler in the setting of integer partitions, the nature of the resulting series boils down to relatively simple pairing of partitions in a sign-reversing manner, showing that the series expansion has coefficients from {−1, 0, 1}.The proof is made even simpler by the geometric representation of partitions using Ferrers' diagrams, which makes it clear which partitions cannot be paired, and provides an explicit formula for the exponents of the terms with ±1 coefficients.This of course is the beautiful pentagonal number theorem of Euler [8], and Fabian Franklin's combinatorial proof of this result [10].
The reciprocal of Euler's product is the generating function of unrestricted integer partitions, and the above determination of the coefficients gives the surprising recursion for the partition function where the indices decrease by generalized pentagonal numbers.
The starting point of the present paper is the consideration of the analogous product where the exponents run through the sequence of Fibonacci numbers (F n ) n≥2 : 1, 2, 3, 5, 8, 13, . . .instead.
The main result of the paper is that the coefficients of the expansion of (2) are also from {−1, 0, 1} (Theorem 9.1).They can be efficiently calculated, and straightforward formulas can be obtained when various interesting sequences of integers are taken as exponents.
There are some surprising properties of the expansion in (2).For one thing, the indices of the so-called canonical Fibonacci representation of an exponent n play a central role in the calculation of the corresponding expansion coefficient in the development of (2).The canonical representation of n (also called the Zeckendorf representation) is the unique representation n = F k 1 + F k 2 + • • • + F k r where k j − k j+1 ≥ 2 for j = 1, 2, . . ., r − 1 and k r ≥ 2. The partition associated to n is then with parts differing by at least two, and the smallest part ≥ 2. For the analysis of the coefficients of the expansion of (2) the parts can be taken to be remainders k i modulo 4; we show that the coefficient of y n in the expansion only depends on the vector x = (x 1 , x 2 , . . ., x r ) of these remainders (Theorem 8.1).Therefore, the expansion coefficients are determined by such vectors x, or equivalently strings x = x 1 x 2 • • • x r with x i ∈ {0, 1, 2, 3}.Also, there are conditions in terms of forbidden subwords of the string x which guarantee that the associated expansion coefficient vanishes (Theorem 13.2).
It is possible to compute the value of the coefficient of y n in (2) explicitly in various special cases.For instance, if n has canonical representation n = F k 1 + F k 2 +• • •+ F k r with r ≥ 1, then closed form expressions can be found for r = 1, 2, 3 as given in (21),( 25) and (30), respectively.Another type of result gives that if for some fixed p ∈ {0, 1, 2, 3}, k i ≡ p mod 4 for all i, then the coefficient of y n is given by 1, − p 2 , p 2 − 1, depending on whether r mod 3 is 0, 1, or 2, respectively (Theorem 12.1).Another result of this type is that the coefficient vanishes whenever r ≥ 2 and k 1 − k 2 ≡ 3 mod 4 (Corollary 11.1).
There are other combinatorial aspects of the expansion (2).For example, the coefficients ϑ(x) over all x (i.e., all positive integers) are completely determined by a monoid of 25 matrices M 1 , M 2 . . ., M 25 , each 2 × 2 matrix determining a formula for the coefficients depending on the canonical representation of the exponent n (Theorems 11.1, 13.1).
The monoid associated with these 25 matrices naturally defines a finite Markov chain, and assuming that the elements of the vector of residues x are picked independently and uniformly, asymptotic probabilities for the values of the coefficients can be determined as a function of the number of summands r of the canonical representation of the exponent (Theorem 17.1).

Related Work
The interest in this subject was initiated by the work of Carlitz, especially by his studies on Fibonacci representations [3,4], and the properties of the product which is the expansion (2) studied here but with plus signs.We make use of his idea of adding an auxiliary variable to the generating function to derive the central recursions.Carlitz also proved a curious result on a property of Fibonacci representations that we make use of repeatedly.
In addition to Carlitz, Klarner [14,15] also studied the number of representations of an integer in terms of Fibonacci numbers, compiling tables of data on these quantities.Earlier work on Fibonacci representations can be found in Daykin [5], Hoggatt [11], and Ferns [9], among many others.In fact, the fascination with the pretty numerical properties of the Fibonacci sequence has produced a wealth of results which are too numerous to be listed here.

Preliminaries
The well-known Fibonacci sequence is defined by A monoid is a set with an associative operation having an identity element.The free monoid of a set is the set * of all finite sequences of zero or more elements of .We will also denote * by and refer to it as the monoid generated by .The elements of * are called words or strings.The unique string in * with no elements is denoted by .A string u is a factor or a subword of w if there exists strings x, y ∈ * with w = xuy.
A partition λ of a positive integer n is a weakly decreasing sequence of positive integers A finite Markov chain is specified by a number of states s 1 , s 2 , . . ., s m , and a sequence of steps through these states so that when the process is in state s i , there is a probability p i j that in the next step, it will be in state s j .The m × m matrix P = [p i j ] is called the transition matrix of the chain.Its entries are nonnegative with each rowsum 1.To specify the process completely, we provide P and a starting state.The probability of moving from state s i to s j in k steps is the i jth entry p (k) i j of the matrix P k .A set of states communicate with each other if the process can move from any state to any other in the set.If some power of the transition matrix has all positive entries, then the chain is called regular.A state is absorbing if once entered, it cannot be left.A chain is an absorbing chain if it has at least one absorbing state and if it is possible to reach an absorbing state (possibly in many steps) from any state.We refer the reader to [13] for details.

Euler's Expansion
Our starting point is the famous product of Euler [8].A wonderful account of this product and its impact can be found in Andrews [2].
The combinatorial interpretation of the coefficients of the power series expansion of (3) and Franklin's proof of it is a favorite topic in any introduction to combinatorics course and can be found in many sources in the literature (see for example [1], [16], [7]).Computing a few terms of the product, we obtain the following expansion.
As is well known, the reciprocal of (3) is the generating function of the number p(n) of unrestricted partitions, which in view of (4) gives Euler's surprising recursion for it.

The Main Product
In analogy with Euler's product where the exponents are 1, 2, 3, 4, . .., we consider the infinite product where the exponents run through the Fibonacci numbers (F n ) n≥2 : 1, 2, 3, 5, 8, 13, . . .A few terms of this expansion are Evidently the reciprocal of (5) is the generating function of integer partitions into parts that are Fibonacci numbers F 2 , F 3 , F 4 , . ... Note that the Fibonacci number F 1 = 1 is not used in (5) to avoid having two different kinds of 1, as we already have F 2 = 1.Coefficients of the powers y n in the expansion for the first few values are as shown in Table 1.There seems to be no simple periodic behavior of the location of the ±1 coefficients.We let v(n) denote the coefficient of y n in the series expansion in ( 5) and refer to it as an expansion coefficient.Reading from the list in Table 1, we have etc., with this notation.

Fibonacci Representations
Every positive integer n can be written in terms of Fibonacci numbers as a sum where and r depends on n.We call (6) a Fibonacci representation, or simply a representation of n.By a theorem of Zeckendorf [17], the representation ( 6) is unique if we impose the conditions: This unique Fibonacci representation of n will be called its canonical representation (it is also referred to as the Zeckendorf representation in the literature).To aid in our calculations, the following notation will be useful.Given a vector of indices k satisfying (7), we let but this is not canonical because of the presence of F 1 , and F 9 + F 6 + F 3 + F 2 is not canonical either because of the presence of consecutive indices.We write 45 = n (9,6,4).
Example 2 There are a total of five Fibonacci representations of n = 106 as given below, the first being the canonical representation.
The first and the second columns list the representations of n into an even and an odd number of Fibonacci summands, respectively.The corresponding expansion coefficients v(n) of the product ( 5) are given in the last column It is also common to use binary strings to encode such representations by using their Fibonacci indicator "digits," where the rightmost digit corresponds to F 2 .For the above five representations of 106, these strings are 1000100101, 1000011101, 110100101, 110011101, and 101111101, respectively.The encoding of a canonical representation is called a Fibonacci string.Fibonacci strings of a given length form the vertices of a Fibonacci cube.Fibonacci cubes form a family of networks which has many interesting combinatorial properties [6,12].
Table 2 is a list of all Fibonacci representations of the integers 1, 2, . . ., 12.As examples, for n = 3, there is a single representation with an even and a single representation with an odd number of terms, and therefore the coefficient of y 3 in ( 5) is 0. For n = 8, there are two representations with an odd number of terms and one representation with an even number of terms, so the coefficient of y 8 in ( 5) is −1.
If n has a Fibonacci representation as given in ( 6), then we use the following notation due to Carlitz: with e(0) = 0.
Carlitz [3] proved the following interesting result, which justifies the use of an arbitrary Fibonacci representation in (8).
Theorem 5.1 (Carlitz) The value e(n) is independent of the Fibonacci representation in ( 6) chosen for n.
Following along the lines of an idea by Carlitz, let Equating coefficients in the resulting identity where α(0, 0) = 1, and α(m, n) = 0 if either argument is negative.Note that the expansion of the product ( 9) can be written in the form: 1 over all r ≥ 1.Therefore, if the exponent of y in a summand in (11) above is n, then the exponent of the corresponding x of that term is e(n).
Remark 1 For any given n, α(m, n) = 0 unless m = e(n).We will use this observation repeatedly.
Using Carlitz's theorem, we can write in which the second equality is a consequence of (10) with m = e(n).
We prove a sequence of three lemmas and collect the partial results together as Proposition 6.1.
Proof In view of (13), it suffices to show that n−e(n) = e(e(n)) and α(n−e(n), e(n)− 1) = 0. We have To prove that α(n −e(n), e(n)−1) = 0, we start with the standard Fibonacci identities Identity (15) implies that Since this is a Fibonacci representation, Carlitz's theorem with identity (16) imply that Therefore, the first argument in α(n − e(n), e(n) − 1) is not equal to e(e(n) − 1) and it must vanish.
The following useful lemma is another consequence of the pair of Fibonacci identities in (15) and (16).
Next, we consider the case in which k r is even.
We collect these results in Proposition 6.1.

Proposition 6.1 Let n have the canonical Fibonacci representation n
where n 1 and e is the operator defined in (8).

A Special Case: r = 1
It is possible to we give an explicit formula for the expansion coefficients for r = 1, i.e., when the exponent n = F k 1 is a Fibonacci number.
Remark 2 Proposition 6.1 together with Lemma 6.4 provide a recursive algorithm to compute v(n).As an example, for n = 18 = F 7 + F 5 , However, this approach to compute v(n) is not very illuminating.
Suppose that e is defined as in (8).We recursively define its powers for t ≥ 0 by setting e t (n) = e(e t−1 (n)) with e 0 (n) = n.Additionally, given a vector of indices k satisfying (7), we let

A Lemma on Staircases
We need a result on the expansion coefficients where the canonical representation of n ends with a "staircase."This result is used in the proof of the main recursion for the coefficients, given as Proposition 7.1 in the next section.Lemma 6.5 For s ≥ 1, ϑ(k, 2s, 2s − 2, . . ., 4, 2) = (−1) s ϑ(E 2s−1 (k)) .

By induction on s,
and the lemma follows.
then the conclusion is the third case of Proposition 6.1.If 2t > 2, then the second case of Proposition 6.1 and the expression (19) for e(n) − 1 imply that where the last equality is a consequence of the first part of Proposition 6.1.If 2t −2 > 2, we can apply to same expansion to the first term above to obtain ϑ(e 2 (k), 2t − 2) = ϑ(e 3 (k), 2t − 3) + (−1) t−1 ϑ(e 2(t−1)−3 (e 3 (k))) and therefore, Continuing this way, there is a cancelation for t odd; therefore, for t odd, we obtain Using the first and the third parts of Proposition 6.1, this gives for k r = 2t, t odd.When k r = 2t and t is even, For k r = 2t + 1, the first part of Proposition 6.1 implies that and the result follows from the k r = 2t part of the proof with e(k) playing the part of k.
The different cases in Proposition 7.1 depend on the value of k r modulo 4. They can be combined to write the recursion for v(n) in the form There is even a simpler formulation if we allow ourselves some further notation by setting x i = k i mod 4 for every index i.Using also the notation used in Example 1, we record this as a proposition in the following form.

Invariance Modulo 4
A surprising application of Proposition 7.2 is the following result.

Theorem 8.1 Suppose n has the canonical representation n
Proof This is clearly so if r = 1 because of the formula (21) for this case.The formula given in Proposition 7.2 for r > 1 depends on x r ≡ k r mod 4, and values for numbers represented by E k r −2 (k) and E k r −1 (k).These numbers have k 1 values in their canonical representation that are smaller than that of n, and the proof follows by induction.

Remark 3 This result means that if n
Since v(n) only depends on the vector of remainders we denote v(n) by ϑ(x 1 , x 2 , . . ., x r ) or equivalently by ϑ(x).Similarly, we denote n by n(x).Of course, given a vector x = (x 1 , x 2 , . . ., x r ) with x i ∈ {0, 1, 2, 3}, there are infinitely many integers n whose canonical representation indices modulo 4 are the given x i , i.e., infinitely many n = n(x).
Remark 4 From now on arithmetic operations involving parameters x i and vector x are understood to be modulo 4 to avoid cumbersome notation.For instance, when we write x 1 − x 2 − 2, we mean (x 1 − x 2 − 2) mod 4. As another example, the notation is a shorthand for (x 2 −x 3 −1) mod 4 2 .

Expansion Coefficients Lie in {−1, 0, 1}
Since the expansion coefficients only depend on the remainder of the indices of the canonical representation modulo 4, the recursion of Proposition 7.2 can be restated as follows.
where x = (x 1 , x 2 , . . ., x r −1 ).This can be split into four recursions depending on the value of x r : Next, we prove that the expansion coefficients lie in {−1, 0, 1}.To do this we strengthen the induction hypothesis and prove the following result.Proof This is the case for r = 1 given in Lemma 6.4.For r > 1, from (24) we have These differences are in {−1, 0, 1}, since the summands in each do not have opposite signs and they are in {−1, 0, 1} by the induction hypothesis.This holds for the last two cases in (24) also.To complete the induction proof, we need to show that a = ϑ(x, x r ) and b = ϑ(E(x, x r )) never have opposite signs, i.e., ab = −1.For x r = 0, we have a = ϑ(x, 0) and b = ϑ(E(x, 0)) = ϑ(x − 1, 3).From ( 24) we have ab = ϑ(E 2 (x)) 2 = −1.The proofs of the other cases are similar.
10 Another Special Case: r = 2 When r = 2, n is equal to the sum of two nonzero, nonadjacent Fibonacci numbers.Proposition 10.1 For r = 2, the expansion coefficients are given by Proof The formula follows from ( 23) and ( 21), keeping in mind that the operations involving the x i are performed modulo 4.
Remark 5 Note that for r = 2, the expansion coefficients for x 1 − x 2 = 3 are all zero as a consequence of (25).We will prove (Corollary 11.1) that this holds for arbitrary r ≥ 2.

The Matrix Recursion
When we have a linear homogeneous recursion, it is usually helpful to formulate it as a matrix recursion.For constant coefficients, we then use linear algebra techniques to compute the powers of the matrix and obtain Binet-like formulas for the n-th term of the sequence.When the coefficients are not constant, the matrices involved need to have special properties if we are to have any hope of writing the resulting matrix product in a simple form.For 2 ≤ i ≤ r − 1, define the 2 × 2 integer matrices where the arithmetic operations involving the x i 's are done modulo 4 as before.
where the empty product for r = 2 is taken to be the 2 × 2 identity matrix.Then the value of the expansion coefficient ϑ(x, x r ) is given by the vector-matrix-vector product Proof For r = 2, from formula (25) and Lemma 6.4 we obtain the following expression.
Using recursion (23), Therefore, our proof obligation is to show that The above identity is immediately verified using Mathematica to check the equality of both sides for the sixteen possible values of x r −1 and x r modulo 4.
An immediate consequence of Theorem 11.1 is the following corollary.

Corollary 11.1 If n has canonical representation n
the corollary follows from (28).

Formula for r = 3
For = 3, the matrix formulation gives Therefore, the formula for the expansion coefficient ϑ( The formulas ( 21), ( 25) and (30) provide closed form expressions for the expansion coefficients for r = 1, 2, 3. Evidently, these formulas get unwieldy very quickly.
12 The A-Matrices and a Special Case Among all possible 2 × 2 matrices B i , there are only four distinct ones, depending on the value of the difference x i − x i+1 modulo 4. We will refer to these four matrices indexed by the possible remainders 0, 1, 2, 3 as the A-matrices.They are given in (31).
We can use the matrix recursion formulation to obtain formulas for the expansion coefficients explicitly in certain other special cases.Suppose in x = (x 1 , x 2 , . . ., x r ) with r ≥ 3, all digits have the same remainder p modulo 4, for some p ∈ {0, 1, 2, 3}.In this case each A-matrix that is a factor of M(x) is equal to A 0 as given in (31), and the expansion coefficient is given by We calculate that for m ≥ 0, This gives the following result.
Table 3 The 25 elements of the monoid of M-matrices generated by the four A-matrices for k ≥ 0. Proof For r ≥ 3, the proof is immediate from the vector-matrix-vector product in (32) and the expression for the powers of A 0 in (33).We check that the values for r = 1 and r = 2 already obtained for x 1 = p (Lemma 6.4) and x 1 = x 2 = p (Proposition 10.1) are also accounted for by the formula in (34).

The Monoid of M-Matrices
The four A-matrices A 0 , A 1 , A 2 , A 3 defined in (31) generate a monoid where the empty product is taken to be the 2 × 2 identity matrix I .Elements of this monoid coincide with the possible matrices M(x) that appear in (29) for the computation of the expansion coefficients.We will refer to them as the M-matrices.
The set of M-matrices was determined by experimentation with products of the Amatrices in (31) using Mathematica.Once determined, it is straightforward to prove that they are indeed the monoid A 0 , A 1 , A 2 , A 3 .It is interesting that there are exactly 25 M-matrices.These are labeled M 1 through M 25 as shown in Table 3.

Theorem 13.1
The monoid generated by the A-matrices consists of the 25 matrices M 1 through M 25 of Table 3. Incidentally, Table 4 gives a straightforward algorithm to solve the problem of finding the word in the last column to which a given element of the monoid A 0 , A 1 , A 2 , A 3 is equal.For example given the word w

The Exact Distribution of the Expansion Coefficients
Since the values of the expansion coefficients are determined by their residues x = x 1 x 2 • • • x r for an integer n = n(x) with r summands in its canonical representation, we can use the explicit formulas for the r = 1 and r = 2 cases from Lemma 6.4 and Proposition 10.1, and for r > 2, use the vector-matrix-vector formula of (28) to compute the distribution of the −1, 0, 1 values for the expansion coefficients as a function of r .These brute-force computed values are given as Table 5.
If we plot the values of exact values of these probabilities, we see that the values for −1 and 1 expansions oscillate around a common value (Fig. 1).
Let u r denote the empirical probability that the expansion coefficient is nonzero (this is the sum of the green and the blue lines in Fig. 1).To get an idea of how this ratio decreases using the available numerical data, we compute the values of u r +1 /u r for r = 1, 2, . . ., 12.This gives the sequence of ratios The decimal expansion of the last three ratios are about 0.8495, 0.8492, 0.8464.We will show in Theorem 17.1 that both −1 and 1 probabilities go to zero as 2+

M-Matrices as a Finite Markov Chain
We assume that modulo 4 remainders of integers n = n(x) are distributed uniformly and independently.Thus, for a given r , the components x i ∈ {0, 1, 2, 3} of x = (x 1 , x 2 , . . ., x r ) are each picked independently with probability 1  4 .The Markov chain we construct has as its states M 1 through M 25 .There is a transition from state M i to state M j if M i A k = M j according to Table 4, where A k is one of A 0 , A 1 , A 2 , A 3 .The probability of each transition is 1  4 .We refer to the resulting finite Markov chain as the M-chain.The underlying state diagram of the M-chain (without the probabilities labeled) is given in Fig. 2. Since all transitions from M 12 are back to itself, M 12 is absorbing.The process starts at state M 23 .

The Nature of the M-Chain
The M-chain decomposes into strongly connected components (communicating states) which consist of the classes given in (37).
The restriction of the chain to each of the classes is regular in that the transition matrix of the chain restricted to the elements of the class has a matrix power whose elements are all positive.For example, the cubes of the restricted matrices for the classes C 2 , C 3 , C 4 are identical and is given by the matrix on the left in (38); whereas the cube of the restricted transition matrix for C 5 is the one on the right in (38).
The schematic of the patterns of communication between these classes is as shown in Fig. 3. for the number m r , o r z of −1, 1, and coefficients, respectively, as a of r .

Remark 9
The interesting point about these three recursions is that even though they have been obtained from probabilistic considerations by looking at the behavior of the Markov chain of M-matrices, they hold for the actual values computed by brute force and represented in Table 5.There is probably a simpler combinatorial explanation.

Conclusions and Remarks
The main result presented here is that the coefficients of the expansion of (2) are from {−1, 0, 1}, as is the case in the expansion of Euler's product (1).
The coefficients in the development of (2) can be efficiently calculated, and explicit formulas can be obtained when various interesting sequences of integers are taken as exponents.However, a result like the pentagonal number theorem with explicit identification of the ±1 coefficients, and a simple combinatorial proof of by using Ferrers' diagrams with Fibonacci numbers as parts seems difficult.Instead of partitions with distinct parts that occur in the case of the combinatorial interpretation of the expansion of (1), the relevant partitions for this problem seem to be integer partitions formed by the indices of the canonical Fibonacci representation of n.
There are conditions in terms of forbidden subwords of the string x, whose letters are the modulo 4 remainders of the indices of canonical representation of an exponent, which guarantee that the associated expansion coefficient vanishes.
A more extensive formal language oriented study of the properties of the coefficients is made possible by the construction of a deterministic finite automaton over the alphabet = {A 0 , A 1 , A 2 , A 3 } which has as its states the states of the M-chain in Fig. 2, and where the transitions are labeled with the letters of as indicated by right multiplication in Table 4.
Various submonoids of the M-matrices give results on the coefficients of subsequences of exponents.We have considered the case of A 0 in Sect.12.The analysis of this submonoid gives the result in Theorem 12.1 in which the parts in the canonical representation of the exponent are all equivalent to the same remainder p modulo 4. Study of various other submonoids gives results of a similar nature from which asymptotic expressions can also be obtained.The simplification of the analysis in these cases is made possible by making use of the relations satisfied by the A-matrices as discussed briefly in Sect.13.
An example of a result based on the study of submonoids and the uniqueness of the canonical representation is Proposition 19.1, which we state here without proof.
Proposition 19.1 Writing the product in (2)

Fig. 2
Fig.2The Markov chain (M-chain) defined by the matrices M 1 , M 2 , . . ., M 25 as The start state is M 23 ; M 12 is the absorbing state.The probability is1  4 for each transition

Fig. 3
Fig. 3 Sketch of the structure of the M-chain with the communicating classes C 1 , C 2 , C 3 , C 4 C 5 as given in (37)

Table 4
The table of M-matrices and the effect of right multiplication of the M i by the four A-matrices M5For example, M 6 A 0 = M 8 and M 6 A 1 = M 12 .The last column is the lexicographically smallest word (product) in A 0 , A 1 , A 2 , A 3 that is equal to the M i in that row.The null word in row 23 corresponds to the identity matrix I