Minimal n for 6 term Engel Expansion of 6/n

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1 Minimal n for term Engel Expansion of /n Elvia Nidia Gonzalez Summer 0 University of California, Riverside MSRIP University of California, Berkeley UC LEADS Mentor: J. Bergner PhD, UC Riverside Math Department July, 0 - Motivation - Abstract While begining on my summer research project, I began creating tables whose entries represented the reciprocals of integers whose sum add up to fractions of the form n, 4 n,..., x n. I did this in the hopes of observing a pattern for the smallest such x such that an Engel series of length x is produced (an Egyptian fraction decomposition can be obtained directly from the Engel series). Completing the tables and highlighting any desired cases was simple enough with the aide of a computer, but still tedious. Additionally, this brute force method will not yield results for every possible x. In order to acheive this, I need to prove for any n, it is the smallest natural number that produces a series of length x for x n (0, ). In order to improve my techniques in proof writing, I decided to start with a very small case in the hopes that after working through proofs of this case, I will better be able to work out a technique for a general case. A proof that every Engel series is finite for every rational number is included. Introduction Suppose that n (0, ) (so n > must hold). The smallest possible n that will produce a six term Egyptian fraction decomposition using an Engel series is. What is an Engel Series? An Engel series is sequence a, a,..., a n,... [4] which I will write as a multiset {a, a,..., a n...}. Any real number x can be derived using an Engel

2 series in a unique way. This x can be represented as i= How can an Engel series be computed? An Engel series for x n can be computed as follows: a a a i []. u = x n a = u u i+ = u i a i a i+ = u i This algorithm halts when a u i = 0. Each a i forms an element of the multiset {a, a,..., a k }. The Egyptian fraction decomposition is then: x n = a + a a Theorem Engel series are finite for all rational numbers. a a a k Proof Suppose m Q. Then m = x n for x, n Z. Let m = u. Using the Engel series expansion algorithm, u = x n a = n x u = u a = ( x n a ) = x a n n Next, n x = a if and only if a < n x < a. ([], page 9). Thus I have: a < n x x a x < n x a n < x The numerator for u is less than the number for u. This continues and the numerator for u i > u i+. Since each is an integer, eventually u k will have a

3 numerator of 0 hence u k = 0 and the algorithm halts. This will produce a finite multiset {a, a,..., a k } as desired. From here on out, a specific case for x = will be the focus. The intent is to show n = is the least n such that the Egyptian fraction decomposition produced using Engel series is finite. The only possible numbers that need to be checked, that is, eliminated as a possibility for n are 7 through 0. The following three identities from [] will be used throughout this paper: Case : n = p () r + s = r + s for s Z (Page 9) () r s = r+s s (Page 9) () r = r r Z r = r (Page 8) Suppose n = p. Then n = p = p. This is a one-term Engel expansion. Hence the following numbers are eliminated from consideration for n are, 8, 4, 0,, 4, 48, 54, 0. Case : n = p Suppose n = p. Then n = p = p.. Subcase : p is even If p is even, then p then p can be written as p = p for p N. Thus as in case. n = p = p = p = p. Subcase : p is odd If p is odd then p = p +. Now apply Engel s algorithm. u = n = p = p = p + a = p +

4 By () a can be rewritten using the floor function. a = (p + ) + = p + = p + Next, I can use () to separate the from the floor function. a = p + Finally, since p N Z by () p = p and a = p +. Find u by continuing the algorithm for Engel series. u = a u = (p + ) To find a take the ceiling of the reciprocal of u. p + = p + a = p + = p + = p + Iterating again to find u = a u = p + (p + ) = 0 and the algorithm halts. Hence the Engel expansion is n = p = (p + ) = p + = p + + (p + )(p + ) Which is still not a six-term expansion. Now eliminated as candidates for n are 9, 5,, 7,, 9, 45, 5, and 57 as well as those mentioned at the end of case. 4 Case : n = p Suppose n = p. Then n = p = p. I have 8, 0, 4,, 0,,, 8,, 4, 8, 40, 44, 4, 50, 5, 5, and 58 to eliminate. 4. Subcase : p is a multiple of If p is a multiple of, I can write p = p for p N and n = which has already been covered by case. p = p 4

5 4. Subcase : p mod If p mod then p = p + and n = p = ( p + ) = p + = u a = p + = (p + ) + = p + 4 = p + 4 Since p N Z, p can be removed the floor and a can be rewritten as: a = 4 + p = p + Next, continue by finding u and a. u = u a = p + (p + ) = p + a = p + = p + = p + Finally, u = u a = p + (p +) = 0 and the algorithm halts. The Engel expansion for this case becomes: n = (p + ) = p + + (p + ) (p + ) Which is less than six terms long. Now eliminated as possibilities for n are 0,,, 8, 4, 40, 4, 5, and Subcase : p mod Notice that if p mod then p = p +. The following two cases will explore the possible Egyptian fraction decomposition lengths whenever p is odd or even. 5

6 4.. Sub-Subcase A: p is odd If p is odd then p = p + for some p N and u = n = p = p = p + a = p + = (p + ) + u = = p + = p + = p + = p + p + (p + ) = p + a = p + I know that p is odd and positive so I can rewrite it as p = p + (again p N). I can now continue to rewrite a. a = (p + ) + = p + 4 = p + = p + = p + Since p = p +, clearly p = p and I can rewrite a one last time in terms of p. a = p + = ( p + ) + = p + The next iteration will produce u i = = 0 and the algorithm halts. For this subcase the Engel series for the Egyptian fraction decomposition is: p + p + n = p = p = p + + (p + ) (p + ) This is still not a six term Egyptian fraction decomposition using Engel series.

7 4.. Sub-Subcase B: p is even Start by applying the algorithm for an Engel series. As in 4.., both u anda are the same. u = p + a = p + = p + Now continue by finding u and a. u = p + (p + ) = p + p p = + a = p + p + In order to simplify a further, I will use the fact that p is even so it can be rewritten as p = p for some p N. a = p + = (p + ) + Since p = p, p = p so I can rewrite a. a = p + = p + Then continue with the algorithm. u = p + p + = a = u = p + Finally u 4 = p + p + = 0 and we halt. = p + = p + = p + = p + p + This produces a three term Engel series of {p +, p +, p + } and the Egyptian fraction decomposition is : n = p = p = p + + (p + ) (p + ) + (p + ) (p + ) (p + ) This leaves only 7,,, 7, 9,, 5, 9,, 5, 7, 4, 4, 47, 49, 5, 55 & 59 as possible choices for n. 7

8 5 The remaining n < 0 5. n 5 mod Since n mod, n = p + 5 for some p N. Applying the algorithm for Engel sereis produces the following: u = n = p + 5 a = p + 5 (p + 5) + = = p + 0 = 0 + p = p + u = (p + ) = p + 5 p + 5 a = p + 5 = p + 5 = p + 5 Finally, u = p+5 (p+5) = 0. This produces the following Egyptian fraction decomposition: 5 = p + + (p + )(p + 5) which is still not the desired length. 5. n mod I have yet to finish this part Conclusions & Future Work Clearly then, is the smallest n such that n has a six term Engel expansion. This poses other questions. Which n is the least n such that 7 n has a seven term Engel expansion? What about for 8 n and so on? What about a general case? That is, which n produce and x term Engel expansion for x n. 8

9 References [] Erdos, Shallit. New bounds on the length of finite pierce and Engel Series. Jornal de theorie des nombres de Bordeaux.. (99): 4-5 [] Graham, Knuth, Patashink. Concrete Mathematics. Addison Wesley, Massachusetts, st Edition, 989. [] V. Laohakosol, T. Chaichana, J. Rattanamoong, and N.R. Kanasri. Engel Series and Cohen-Egyptian Fraction Expansions. International Journal of Mathematics and Mathematical Sciences (009): 5pages. [4] Renyi, A. A New Approach to the Theory of Engel s Series. Annales Universitatis Scientiarum Volume 5 (9): 5-. 9