Families of Rational Numbers with Predictable Engel Product Expansions

R. S. Melham Department of Mathematical Sciences University of Technology, Sydney PO Box 123, Broadway, NSW 2007 Australia

                     ray.melham@uts.edu.au

Abstract In 1987 Knopfmacher and Knopfmacher published new infinite product expansions for real numbers 0 < A < 1 and A > 1. They called these Engel product expansions. At that time they had difficulty finding rational 0 < A < 1 for which the Engel product expansion is predictable. Later, in 1993, Arnold Knopfmacher presented many such families of rationals. In this paper we add to Arnold Knopfmacher’s list of such families.

1 Introduction

Knopfmacher and Knopfmacher [1] proved that every positive real number 0 < A < 1 has an expansion of the form

A = ∏_{n=1}^{∞} (1 − \frac{1}{a_{1}a_{2} ⋯ a_{n}} ) ,  (1)

where a_{n} is a positive integer for n ≥ 1, a_{1} ≥ 2, a_{n+1} ≥ a_{n} −1 for n ≥ 1, and a_{n} ≥ 2 infinitely often.

These representations were called Engel product expansions. To abbreviate (1), Knopfmacher and Knopfmacher wrote

A = ((a_{1}, a_{2}, a_{3}, …)).

They remarked that there do not appear to be simple cases where the digits {a_{i}} could be found with explicit formulae. Later, Arnold Knopfmacher [2] gave several families of rationals where, from some point onwards, the digits {a_{i}} in the expansion (1) satisfy the recurrence

a_{n+1} = (a_{n} + 1) a_{n−1} − 1.  (2)

To illustrate, one of Knopfmacher’s main results is the following:

Theorem 1. Let a and m be positive integers with a ≥ 3 if m = 1, and a ≥ 2 otherwise. Then

\frac{(a − 1) m − 1}{am − 1} = ((a_{1}, a_{2}, a_{3}, …)) ,

where a_{1} = a, a_{2} = (a − 1)m, a_{3} = am − 1, a_{4} = a_{1}a_{2} − 1, and (2) applies for n ≥ 4.

Knopfmacher and Knopfmacher [1] showed that the expansion (1) is unique if it has non-decreasing digits. Therefore, in Theorem 1 a unique expansion is guaranteed for every a provided m ≥ 2.

Knopfmacher [2, page 426] stated that it would be interesting to characterize in some way which rational numbers have expansions where the digits {a_{i}} ultimately satisfy the nonlinear recurrence (2). Based on our investigations it seems difficult to resolve this completely. We have, however, managed to add to Knopfmacher’s collection of such rationals. We present our results as families of rationals that require two or more parameters to describe them. Indeed, we have found many one-parameter families, but, to conserve space, we do not present them here.

In Section 2 we state and prove one of our main results. In Sections 3 and 4 we state further results.

2 A Result With Sample Proof

Before proceeding we give the greedy-type algorithm from [1] which was used to derive the expansion (1): Given any 0 < A < 1, let A_{1} = A, a_{1} = 1 + ⌊\frac{1}{1−A_{1}} ⌋, where ⌊x⌋ denotes the floor of the real number x. Then recursively define, for n ≥ 2,

a_{n} = 1 + ⌊\frac{1}{(1 − A_{n}) a_{1}a_{2} ⋯ a_{n−1}} ⌋ ,

where

A_{n+1} = \frac{a_{1}a_{2} ⋯ a_{n}}{a_{1}a_{2} ⋯ a_{n} − 1}A_{n}.

The theorem that follows is similar in spirit to Theorem 1 above. The proof we present proceeds along the same lines as Knopfmacher’s beautiful proof of Theorem 1.

Theorem 2. Let a and m be positive integers with a ≥ 2 if m = 1, and a ≥ 1 otherwise. Then

\frac{(2a − 1) m − 1}{4am − 2} = ((a_{1}, a_{2}, a_{3}, …)) ,

where a_{1} = 2, a_{2} = a, a_{3} = (2a − 1)m, a_{4} = 2am − 1 , a_{5} = a_{1}a_{2}a_{3} − 1, and (2) applies for n ≥ 5.

Proof. With the use of the algorithm above we write down A_{1}, a_{1}, A_{2}, a_{2}, A_{3}, a_{3}, A_{4}, and a_{4}. We present each A_{i} in a manner that makes it easy to find a_{i}.

A_{1} = \frac{(2a − 1)m − 1}{4am − 2} = 1 − \frac{(2a + 1)m − 1}{2(2am − 1)},

a_{1} = 1 + ⌊\frac{1}{1 − A_{1}} ⌋ = 1 + ⌊\frac{2(2am − 1)}{(2a + 1)m − 1}⌋ = 1 + ⌊2 − \frac{2m}{(2a + 1)m − 1}⌋ = 2,

A_{2} = \frac{a_{1}}{a_{1} − 1}A_{1} = 2 (1 − \frac{(2a + 1)m − 1}{2(2am − 1)}) = 1 − \frac{m}{2am − 1},

a_{2} = 1 + ⌊\frac{1}{(1 − A_{2}) a_{1}} ⌋ = 1 + ⌊\frac{2am − 1}{2m}⌋ = 1 + ⌊a − \frac{1}{2m}⌋ = a,

A_{3} = \frac{a_{1}a_{2}}{a_{1}a_{2} − 1}A_{2} = \frac{2a}{2a − 1} (1 − \frac{m}{2am − 1}) = 1 − \frac{1}{(2a − 1)(2am − 1)},

a_{3} = 1 + ⌊\frac{1}{(1 − A_{3}) a_{1}a_{2}} ⌋ = 1 + ⌊\frac{(2a − 1)(2am − 1)}{2a}⌋
= 1 + ⌊(2a − 1)m − \frac{2a − 1}{2a}⌋ = (2a − 1)m,

A_{4} = \frac{a_{1}a_{2}a_{3}}{a_{1}a_{2}a_{3} − 1}A_{3} = \frac{2a(2a − 1)m}{2a(2a − 1)m − 1} (1 − \frac{1}{(2a − 1)(2am − 1)})
= 1 − \frac{1}{(2am − 1)(2a(2a − 1)m − 1)},

a_{4} = 1 + ⌊\frac{1}{(1 − A_{4}) a_{1}a_{2}a_{3}} ⌋ = 1 + ⌊\frac{(2am − 1)(2a(2a − 1)m − 1)}{2a(2a − 1)m}⌋
= 1 + ⌊2am − 1 − \frac{2am − 1}{2a(2a − 1)m}⌋ = 2am − 1.

We define a sequence {b_{n}} as follows:

(b_{1}, b_{2}, b_{3}, b_{4}) = (2, a, (2a − 1)m, 2am − 1) , b_{n+2} = b_{1} ⋯ b_{n} − 1, n ≥ 3.

We prove the following assertions for n ≥ 4 with the use of induction.

(i) A_{n} = 1 − \frac{1}{b_{n}b_{n+1}} (ii) a_{n} = b_{n} (iii) b_{n+2} = (b_{n+1} + 1) b_{n} − 1.

Firstly, for n = 4

(i) A_{4} = 1 − \frac{1}{a_{4}(a_{1}a_{2}a_{3} − 1)} = 1 − \frac{1}{b_{4}(b_{1}b_{2}b_{3} − 1)} = \frac{1}{b_{4}b_{5}} ;

(ii) a_{4} = b_{4} by the definition of b_{4}; (iii) b_{6} = (b_{1}b_{2}b_{3})b_{4} − 1 = (b_{5} + 1)b_{4} − 1.

Next we show that the validity of each assertion for n ≥ 4 implies its validity for n + 1.

A_{n+1} = \frac{a_{1} ⋯ a_{n}}{a_{1} ⋯ a_{n} − 1}A_{n} = (1 + \frac{1}{b_{n+2}} ) (1 − \frac{1}{b_{n}b_{n+1}} )
= 1 − \frac{b_{n+2} − b_{n}b_{n+1} + 1}{b_{n}b_{n+1}b_{n+2}} = 1 − \frac{1}{b_{n+1}b_{n+2}} ,

where we have used (iii) in the last step. This proves (i).

a_{n+1} = 1 + ⌊\frac{1}{(1 − A_{n+1}) a_{1} ⋯ a_{n}} ⌋ = 1 + ⌊\frac{b_{n+1}b_{n+2}}{b_{n+2} + 1} ⌋
= 1 + ⌊b_{n+1} − \frac{b_{n+1}}{b_{n+2} + 1}⌋ = b_{n+1},

since b_{n} ≥ 0 for all n. This proves (ii).

b_{n+3} = (b_{1} ⋯ b_{n})b_{n+1} − 1 = (b_{n+2} + 1)b_{n+1} − 1.

This establishes (iii), and the proof is complete.

Note that, according to the discussion in the paragraph that follows the statement of Theorem 1, a unique expansion is guaranteed in Theorem 2 provided a ≥ 2.

3 Miscellaneous Results

In this section, and the next, we state our results without proof since all proofs are similar to that shown above. Knopfmacher [2] took 1^{a} to denote a consecutive occurrences of the digit 1, and we do the same.

Knopfmacher gave the following theorem with m = 1. Theorem 3. Let m be a positive integer, and let a ≥ 1 and b ≥ 0 be integers. Then

\frac{m}{2^{b} (2^{a}m + 1)} = ((2, 1^{a+b−1}, 2^{a−1}m + 1, 2^{a}m + 1, 2^{a}m + 1, a_{3}, a_{4}, a_{5}, …)) ,

where a_{1} = a_{2} = 2^{a}m + 1, and (2) applies for n ≥ 2.

Due to the presence of a^{2}, the following theorem differs in nature from those above. Theorem 4. Let a and m be positive integers with a ≥ 2 and m ≥ 1. Then

\frac{(a − 1)am − 1}{a^{2}m} = ((a_{1}, a_{2}, a_{3}, …)) ,

where a_{1} = a, a_{2} = (a − 1)m + 1, a_{3} = (a − 1)m, a_{4} = a_{1}a_{2} − 1, and (2) applies for n ≥ 4.

In the next three theorems a_{1} and a_{2} immediately precede a_{3}, and (2) applies for n ≥ 2. Theorem 5. Let a ≥ 1 and k be integers. Then the expansion for \frac{m − 1}{2^{a}m} is

((2, 1^{a−1}, k + 1, k, 2k + 1, a_{3}, a_{4}, a_{5}, …)) , m = 2k, k ≥ 1;

((2, 1^{a−1}, k + 1, 2k + 1, 2k + 1, a_{3}, a_{4}, a_{5}, …)) , m = 2k + 1, k ≥ 1;

Theorem 6. Let a ≥ 3 and k be integers. Then the expansion for \frac{3m − 1}{2^{a}m} is ((2, 1^{a−3}, 2, 3k + 1, 3k, 12k + 3, a_{3}, a_{4}, a_{5}, …)) , m = 4k, k ≥ 1;

((2, 1^{a−3}, 2, 3k + 1, 12k + 3, 12k + 3, a_{3}, a_{4}, a_{5}, …)) , m = 4k + 1, k ≥ 0;
((2, 1^{a−3}, 2, 3k + 2, 6k + 3, 12k + 7, a_{3}, a_{4}, a_{5}, …)) , m = 4k + 2, k ≥ 0;
((2, 1^{a−3}, 2, 3k + 3, 4k + 3, 12k + 11, a_{3}, a_{4}, a_{5}, …)) , m = 4k + 3, k ≥ 0;

Theorem 7. Let a ≥ 5 and k be integers. Then the expansion for \frac{9m − 1}{2^{a}m} is ((2, 1^{a−5}, 2, 1, 9k + 1, 9k, 36k + 3, a_{3}, a_{4}, a_{5}, …)) , m = 4k, k ≥ 1;

((2, 1^{a−5}, 2, 1, 9k + 3, 12k + 3, 36k + 11, a_{3}, a_{4}, a_{5}, …)) , m = 4k + 1, k ≥ 0;

((2, 1^{a−5}, 2, 1, 9k + 5, 18k + 9, 36k + 19, a_{3}, a_{4}, a_{5}, …)) , m = 4k + 2, k ≥ 0;

((2, 1^{a−5}, 2, 1, 9k + 7, 36k + 27, 36k + 27, a_{3}, a_{4}, a_{5}, …)) , m = 4k + 3, k ≥ 0;

4 A Host of Results of a Particular Type

Knopfmacher [2], in Theorem 3(iii), gave an expansion for \frac{1}{2^{a} (2^{n} + 1)} when a ≥ 0 and

n ≥ 1. Inspired by this, we decided to search for triples of positive integers (b, c, d) such that

the expansion for \frac{1}{2^{a} (2^{bn+c} + 2d + 1)} is predictable for all integers a ≥ 0 and n ≥ 0. We

restricted our search to 1 ≤ d ≤ 50, first fixing d and then searching for appropriate pairs (b, c). Some d are associated with many pairs (b, c). For example, associated with d = 48 are the following eight pairs (b, c): (48, 24), (48, 27), (240, 121), (336, 172), (336, 173), (528, 270), (528, 303), and (1200, 602). On the other hand, for some d we could find no pair (b, c).

In total we found one-hundred and five triples as described in the previous paragraph. For seventy-nine of these triples (b, c, d) we found that the expansion has the form

\frac{1}{2^{a} (2^{bn+c} + 2d + 1)} = ((2, 1^{a+bn+c−1}, a_{1}, a_{2}, a_{3}, …)) ,  (3)

in which

a_{1} = (2^{bn+c−1} + 2d + 1 + e) / (2d + 1) ,
a_{2} = (2^{bn+c} + 2d + 1) / (2d + 1 + 2e) ,
a_{3} = 2a_{1} − 1,

and (2) applies for n ≥ 3. Here e is a relatively small positive integer that depends upon the triple (b, c, d), and can be found by observation.

In twenty-four of our triples, b = 2c. In these cases it is convenient to write the triples as (2b, b, d). In all such cases we found that

\frac{1}{2^{a} (2^{2bn+b} + 2d + 1)} = ((2, 1^{a+2bn+b−1}, a_{1}, a_{2}, a_{3}, a_{4}, …)) ,  (4)

in which

a_{1} = (2^{2bn+b−1} + d + 1) / (2d + 1) ,
a_{2} = 2^{2bn+b} + 1,
a_{3} = a_{2} + 2d,
a_{4} = 2a_{1}a_{2} − 1,

and (2) applies for n ≥ 4.

The reader can now write down the expansion for \frac{1}{2^{a} (2^{2bn+b} + 2d + 1)} corresponding to

the first of the eight triples mentioned above. For this triple (4) applies. For the remaining seven triples (3) applies, and the values of e are, respectively, 4, 1, 8, 16, 32, 9, and 2. The reader can now use (3) to write down the associated expansions.

Interestingly, we found two triples not of the form (2b, b, d), namely (16, 4, 8) and (16, 12, 8), but where the associated expansions are found with the use of (4). For these two triples we use (4) and replace each occurrence of 2b by 16.

Some further examples of triples (2b, b, d) where the associated expansions are found with the use of (4) are (2, 1, 1), (4, 2, 2), (6, 3, 4), (10, 5, 5), (12, 6, 6), and (18, 9, 9). Further examples of triples (b, c, d) where the associated expansions are found with the use of (3) are (12, 7, 2), (12, 8, 2), (12, 9, 6), (24, 14, 8), (40, 23, 8), and (40, 24, 8). Here, for these six triples, the corresponding values of e are, respectively, 1, 2, 4, 2, 4, and 8.

References

[1] A. Knopfmacher and J. Knopfmacher, A new infinite product representation for real numbers, Monatshefte Math. 104 (1987), 29–44.

[2] A. Knopfmacher, Rational numbers with predictable Engel product expansions, in G. E. Bergum, A. N. Philippou, and A. F. Horadam, eds., Applications of Fibonacci Numbers 5, Kluwer Academic Publishers, 1993, pp. 421–427.

2000 Mathematics Subject Classification: Primary 11A67. Keywords: Engel product expansion, Engel expansion.

Received July 9 2009; revised version received August 5 2009. Published in Journal of Integer Sequences August 30 2009.

Return to Journal of Integer Sequences home page.