On a Diophantine equation of Erdős and Graham

Abstract We study solvability of the Diophantine equation n 2 n = ∑ i = 1 k a i 2 a i , in integers n , k , a 1 , … , a k satisfying the conditions k ≥ 2 and a i a i + 1 for i = 1 , … , k − 1 . The above Diophantine equation (of polynomial-exponential type) was mentioned in the monograph of Erdős and Graham, where several questions were stated. Some of these questions were already answered by Borwein and Loring. We extend their work and investigate other aspects of Erdős and Graham equation. First of all, we obtain the upper bound for the value a k given in terms of k only. This mean, that with fixed k our equation has only finitely many solutions in n , a 1 , … , a k . Moreover, we construct an infinite set K , such that for each k ∈ K , the considered equation has at least five solutions. As an application of our findings we enumerate all solutions of the equation for k ≤ 8 . Moreover, by applying greedy algorithm, we extend Borwein and Loring calculations and check that for each n ≤ 10 4 there is a value of k such that the considered equation has a solution in integers n + 1 = a 1 a 2 … a k . Based on our numerical calculations we formulate some further questions and conjectures.

We study solvability of the Diophantine equation in integers n, k, a 1 , . . . , a k satisfying the conditions k ≥ 2 and a i < a i+1 for i = 1, . . . , k − 1. The above Diophantine equation (of polynomial-exponential type) was mentioned in the monograph of Erdős and Graham, where several questions were stated. Some of these questions were already answered by Borwein and Loring. We extend their work and investigate other aspects of Erdős and Graham equation. First of all, we obtain the upper bound for the value a k given in terms of k only. This mean, that with fixed k our equation has only finitely many solutions in n, a 1 , . . . , a k . Moreover, we construct an infinite set K, such that for each k ∈ K, the considered equation has at least five solutions. As an application of our findings we enumerate all solutions of the equation for k ≤ 8. Moreover, by applying greedy algorithm, we extend Borwein and Loring calculations and check that for each n ≤ 10 4 there is a value of k such that the considered equation has a solution in integers n + 1 = a 1 < a 2 < . . . <

Introduction
In the very interesting book [2] Erdős and Graham stated many number theoretic problems. Some of them are related to Diophantine equations. At page 63 of this book the authors consider the following non-standard Diophantine equation which can be seen as an equation of polynomial-exponential type. The authors stated some questions concerning this equation. For example, they asked whether for each n ∈ N + there is a solution of (1), i.e., we look for solutions in k, a 1 , . . . , a k ; or whether for each k ∈ N + there is a solution of (1), i.e., we look for solutions in n, a 1 , . . . , a k . Moreover, the considered related problem of representations of real numbers α ∈ (0, 2) in the form These questions were investigated by Borwein and Loring in [1]. In particular, in the cited paper, the authors proved that for each k there is a solution of (1). Moreover, they proposed an algorithm which, for a given n, allows to find (conjecturally finite) representation of n/2 n in the form Conjecture 1]. However, they do not investigate other Diophantine questions related to (1). In particular, we are interested in the following questions, which in the light of findings of Borwein and Luring are quite natural. Question 1.1. What can be said about the number of solutions of (1) when k is fixed? Is it possible to enumerate all solutions of (1) for small values of k? Question 1.2. Is it possible to bound a k in terms of k only?
Let us describe the content of the paper in some details. In Section 2 we offer basic theoretical results concerning the solutions of (1). We first enumerate all solutions of equation (1) for k ≤ 8. However, the most interesting part of this section is the proof of the inequality a k ≤ 2 k+2 + 2k(log 2 k − 1) − 4. This answer Question 1.2 affirmatively.
In particular, for any given k, the considered equation has only finitely many effectively computable solutions in integers n, a 1 , . . . , a k .
In Section 3, by solving certain discrete logarithm problems we construct an infinite set K, such that for each k ∈ K equation (1) has at least five solutions in positive integers n, a 1 , . . . , a k . Moreover, we apply a modification of greedy strategy of Borwein and Loring and prove that for each n ≤ 10 4 equation (1) has a solution in positive integers k, a 1 , . . . , a k . As an application of our approach we construct an infinite set R of rational numbers, such that for each x ∈ R the number x has at least nine representations in the form ∞ i=1 a i /2 a i . Moreover, based on our numerical data we formulate precise conjecture concerning the quantity of a k , i.e., a k ≤ 2(n +k). We prove that our conjecture is true for all n satisfying n ≥ 2 k − k.

Theoretical results
We start with some easy observations related to equation (1).
Proof. Let us suppose that (1) has a solution for n. It is clear that a 1 ≥ n + 1 and thus a i ≥ n + i for each i ∈ {1, . . . , k}. Because the function f (x) = x/2 x is decreasing for x ≥ 1 We immediately get the inequality By solving the resulting inequality we get the upper bound for n in terms of k. Indeed, To prove the second part of our theorem let us suppose that a 1 ≥ n + 4. Thus, a i ≥ n + 3 + i for i = 1, . . . , k and in consequence By solving the resulting inequality with respect to n we have n ≤ 5 · 2 k − 6 7 · 2 k + 1 < 1 and get a contradiction. Thus n + 1 ≤ a 1 ≤ n + 3. The divisibility 2 a k −a k−1 |a k is clear. Indeed, multiplying (1) by 2 a k−1 we see that the number is an integer equal to a k 2 a k −a k−1 and hence 2 a k −a k−1 |a k . Finally, to get the last statement we proceed by induction on j ≥ 1. Let us start with j = 1, so n ≥ 3. If a 1 = n + 1, then a 1 ≥ n + 2 and and so n < 3, a contradiction. Let us now take n ≥ 2 j+2 − j − 1. Since also n ≥ 2 j+1 − j, by the induction hypothesis we get that a i = n + i for 1 ≤ i ≤ j. If a j+1 ≥ n + j + 2, then As a consequence we get n < 2 j+2 − j − 1, a contradiction that completes the induction step.
Remark 2.2. We observed that the necessary condition for solvability of (1) is the inequality n ≤ 2 k+1 − k − 2. This condition can not be improved. Indeed, as was observed by Borwein and Luring, if k is fixed and n = 2 k+1 − k − 2, then we have the equality i.e., equation (1) has a solution a i = n + i, i = 1, . . . , k.
The divisibility property noted in the last part of Theorem 2.1 can be strengthened as follows.
Proof. We first note that for i = k − 1 we get 2 a k −a k−1 ≤ a k by Theorem 2.1(ii). We will proceed by induction on In the induction step we perform the same calculations. The only difference is that we multiply equation (1) by 2 a k−j . As a result we obtain the equality and the result follows.
As an immediate consequence from the above result we get the following.
Using the last part of Theorem 2.1 and Corollary 2.4 we can try to list all possible solutions of equation (1) for small values of k. Indeed, part (i) of Theorem 2.1 provides a bound on values of n to check, while part (ii) gives exact values for starting elements in (a i ) for large n. However, for small values of n, part (ii) of Theorem 2.1 is not effective and we need to apply Corollary 2.4 to further reduce the set of possible solutions by bounding a k depending on a 1 . After some considerable amount of computer calculations (case k = 8 took more than two days of computing time) we provide complete list of solutions of equation (1)  As a consequence of our computations we obtain an explicit family of rational numbers having at least three different representations in the form Corollary 2.6. There are infinitely many values of x ∈ Q such that the number x has at least three representations in the form Proof. From Theorem 2.5 we see that for n = 1 the equation (1) has two solutions for k = 3 and one solution for k = 7. Let (b i ) i∈N + be a sequence of positive integers satisfying 15 ≤ b 1 and b i < b i+1 for i = 1, 2, . . ., and suppose that the number Then we have the representations and from the assumption on the sequence (b i ) i∈N + we know that the presented representations are different. In order to make the value of the sum a rational number.
Remark 2.7. The above result was also obtained by Borwein and Luring using three representations of the number 1 4 . However, based on our computational approach, we will show in Corollary 3.6 below, that in fact there are infinitely many rational numbers with at least nine representations of the form We close this section with the positive answer to Question 1.2.
Theorem 2.8. Let a 1 < . . . < a k be a solution to equation (1). Then we have Proof. It can be verified that the inequality holds for all solutions given in Theorem 2.5. So let us fix k ≥ 8 and suppose that a k > 2n + 2k log 2 k holds for some n ≥ 1. Since a k > k−1 ln 2 and the function f (x) = x k−1 2 x is decreasing for x > k−1 ln 2 we get that . By Corollary 2.4 this implies that (2n + 2k log 2 k) k−1 > 2 2n+2k log 2 k−a 1 ≥ 2 n−3+2k log 2 k since a 1 ≤ n + 3. But then 2n + 2k log 2 k k 2 We will show that this inequality is not satisfied for any n. Indeed, for n = 1 we get: and considered functions are monotonic. Increasing n by one, right hand side of the inequality is multiplied by 2, while left hand side by: 2n + 2 + 2k log 2 k 2n + 2k log 2 k This is a contradiction, so the inequality a k > 2n + 2k log 2 k cannot hold.
We prove that there are infinitely many values of k such that N (k) ≥ 5. More precisely, we have Before we prove the above proposition, we describe the experimental strategy which we used. More precisely, we looked for values of u and corresponding value(s) of k such that there is a positive integer solution n of the equation i.e., we look for integral values of the expression Equivalently, we need to consider the polynomial-exponential congruence Note that if for a given u the congruence (2) has a solution in k, then necessarily k < r := ord 2 u+3 −3 (2), where as usual ord m (a) = min{v ∈ N + : a v ≡ 1 (mod m)}. In particular, if k 0 is a solution for u, then for each t ∈ N the number k = rt + k 0 is also a solution. Congruence (2) can be written as hence one has to resolve a discrete logarithm problem. There are exactly 16 values of u ≤ 120 such that (2) has a solution, see Table 1. Table 1 Solutions for k of (2) for u ≤ 120 together with the value of r = ord 2 u+3 −3 (2). Proof of Proposition 3.1. The idea of the proof is the following. If we write f i (x) = r i x + k i , where k i , r i correspond to ith elements in the table above, then to get values of k such that (1) has at least m solutions, it is enough to find solutions of the system for certain 1 ≤ i 1 < i 2 < . . . < i m ≤ 16. We checked all 4368 combinations of five elements subsets of the set of linear functions {f 1 , . . . , f 15 } and found that in each case, the above system has no solutions. In case of four functions we checked 1820 subsets and found exactly six subsets such that the above system has solutions. The simplest solutions are obtained in the case of the linear Diophantine system 28x 1 + 22 = 4092x 2 + 242 = = 26202761468337430x 3 + 5843993308712118 = = 2535300206192230667655098198606x 4 + 364550281031913286431277811782.
The corresponding linear functions give (all) solutions of (2) for u = 2, 9, 55, 99, respectively. A standard method gives the solution where t ∈ N, and the corresponding common value of k is given by In consequence, for given u ∈ {2, 9, 55, 99} and each t ∈ N we get an integer value of n for k = k(t) together with values of a 1 , . . . , a k given by a i = n +i, i = 1, . . . , k −2, a k−1 = n +k+u, a k = n +k+u +1. Thus for any given k = k(t) we have four solutions of (1). One additional solution for k corresponds to n = 2 k+1 − k − 2 and a i = n + i, i = 1, . . . , k.
We finish our discussion with the following: Conjecture 3.2. Let us put The set U is infinite. Our proof of Proposition 3.1 based on the existence of certain elements in the set U. Thus, one can ask the following Question 3.4. Suppose that Conjecture 3.2 is true. Does Conjecture 3.2 imply Conjecture 3.3?
It seems that the most interesting (and difficult) question concerning equation (1) is to whether, for a given n, there is k ∈ N + such that (1) has a solution. Essentially, this is [1, Conjecture 1]. Unfortunately, we were unable to answer this question in full generality. Borwein and Luring proved that for each n ≤ 10 3 equation (1) has at last one solution. We were able to extend the range of computations and prove the following: Theorem 3.5. For each 2 ≤ n ≤ 10 4 the Diophantine equation (1) has a solution in  variables k, a 1 , . . . , a k satisfying a i = n + 1 and a i ≥ n + i for i = 2, . . . k.
We now describe a computational method which was used to get the above result. More precisely, in order to confirm that equation (1) has a solution, the following "greedy" strategy was applied: assuming that we have found the sequence (a 1 , . . . , a l ) such that a l 2 l , we define a l+1 = j where j 2 j is the first term that "fits", that is we take the smallest j such that and hope that this process ends after a finite number of steps. Naive implementation of the procedure above leads to a very slow algorithm for large n, so we apply a different approach that is a slight modification of Algorithm 2 given in [1].
Let x ∈ Q, 0 < x < 2 and define: k 0 = min{k ≥ 1 : k 2 k < x}. We define a sequence S(x) = (x k 0 , x k 0 +1 , . . .) as We say that the sequence S(x) terminates if x i = 0 for some i ≥ k 0 (and for all subsequent values in the sequence). It is not difficult to see that x i is precisely the numerator of the fraction where s j = 1 if j 2 j appears in the sum when applying the greedy strategy for x (with the exception of s j = 0 if x = j 2 j ) and s j = 0 otherwise. Moreover, if the sequence S(x) terminates, then s j = 1 (i.e. j 2 j appears in the representation of x) if and only if x j+1 = 2x j .
The above algorithm was implemented in Mathematica [3] in the following form: Our observations show that to find a representation of x using the greedy strategy we can calculate S(x) and see if it terminates. This has the advantage of being much faster as the only operations involved are multiplication by 2 and subtraction (of integers if x has power of 2 as the denominator). Moreover it can be easily verified that x i < i + 1 for all i ≥ k 0 and so x i < k if the sequence S(x) terminates after k steps (or equivalently the representation for x has k terms), i.e. the numbers x i are feasible.
First of all, we note that our approach is strong enough to present an improvement of Corollary 2.6. More precisely, we prove that Corollary 3.6. There are infinitely many values of x ∈ Q such that x has at least nine representations in the form Proof. To get the result it is enough to find one rational number x with nine representations. The idea is very simple. Suppose that we have x = k 1 i=1 a 1,i 2 a 1,i and for m ≥ 2 we are able to compute the expansion Thus, the number x will have at least m representations We take x = 8 2 8 = 1 32 and applying our greedy strategy we compute Due to size of the sets {a k j ,i : i = 1, . . . , k j }, j = 2, . . . , 9, we do not present them in full.
By adding the value of the series ∞ i=1 pi+q 2 pi+q = (q+p)2 p −q 2 q (2 p −1) 2 , where p, q ∈ N + are chosen that p + q > 3437088, to found representations, we get the statement of our theorem.  With the data needed to get Theorem 3.5 we observed that the behaviour of k = k(n) and a k = a k (n) behaves quite irregular. For example, from our numerical data we collected the following peak (or jump) values of k (see Table 2).
In Fig. 1 we also present the graph of the function k : N ≥2 n → k(n) ∈ N.
Based on our numerical data we formulate the following.
On Fig. 2 we present the behaviour of a k (n)/2(k + n), where the values of a i = a i (n) come from our greedy algorithm.
On Fig. 3 we also present the graph of the function k(n)/n.  Remark 3.8. First of all let us observe that the value n + k in the lower bound cannot be replaced by nothing greater. Indeed, if n = 2 k+1 − k − 2 then we get the exact value a k = n + k. The upper bound for a k stated in the above conjecture is reasonable. More precisely, it is easy to see that our statement is true under additional assumption n ≥ 2 k − k. Indeed, we have