Problems & Puzzles: Problems

Problem 35. More wrong turns...

"I once asked Andre Weil, the legendary mathematician, and also a superb historian,
why the Egyptians did this"
[to represent fractions as a sum of unit fractions],
said
[Ronald] Graham, "and he said 'They took a wrong turn..."
The Man who loved only numbers, by Paul Hoffman, p. 170

Any Unitary fractions 1/a, a = integer is also called an Egyptian fraction.

One particular interesting equation involving distinct Unitary fractions and widely studied is the following one:

S(1/ai ) = 1, for i =1 to k>1such that if ai = aj then i = j  ...................... [eq. 1]

The sum of every set of ai integers that satisfy  is called an "Egyptian number"

The least particular example of [eq. 1] is:

1 = 1/2+1/3+1/6, that provides the Egyptian number 11 = 2 + 3 + 6.

Is well known that there is not any solution to [eq. 1] if all the ai integers are asked to be:

* Consecutive numbers (Taeisinger, Kurschak)
* Numbers in Arithmetic Progression  (Nagell)
* Prime numbers

Other peculiar numerical results are:

1. John Leech says that if all the ai integers are distinct odd numbers, then you need at least 9 of them to satisfy [eq. 1], being the largest denominator at least 105. For some unknown reason, in the D11 section of the well known  R. K. Guy's book (UPiNT2) is offered only an example composed by 11 odd numbers: 3, 5, 7, 9, 15, 21, 27, 35, 63, 105, 135

2. Barbeau found a solution of [eq. 1] using 101 distinct positive integers no one dividing each other.

3. Allan Johnson found a solution of [eq. 1] using 48 distinct positive integers all of them having two prime factors (*)

***

Regarding the Leech problem & predictions, Jaime Ayala and myself got some weeks ago not one but four solutions to [eq. 1] using only 9 odd values (**) , confirming both of the Leech's predictions.

Here is our 'minimal' solution using 9 distinct odd numbers and having the least Egyptian number:

1/3+1/5+1/7+1/9+1/11+1/15+1/35+1/45+1/231 = 1

Questions:

1) Can you find the other 3 solutions using 9 distinct odd numbers? Is there other of these solutions unknown to us?

2) We think - according to our method - that our solution, shown above, is the least solution to [eq. 1] using only distinct odd numbers, in the double sense that it has only 9 odd numbers and also has the minimal Egyptian number. Is this true?

3) Can you find a solution that improves the Barbeau solution?

4) Can you find a solution that improves the Allan Johnston solution?

Other interesting references:

a) Eric Weisstein World of Mathematics
b) Kevin S. Brown

_____________________
* In the Barbeau and in the Johnson respective problems, the condition of the Leech problem (of being all the numbers odd numbers) does not applies.

** As a matter of fact the August 6, 2000 we informed of these solutions to R. K. Guy who answered the 22th of the same month: "... I am filing ready for UPINT3, should it ever appear. Best wishes, RKG"

Solution

Jud McCranie found (October 5, 2000) the other 3 solutions that we have found and a fifth one more!...

3 5 7 9 11 15 21 135 10395  <--- this is the unknown for us
3 5 7 9 11 15 21 165 693
3 5 7 9 11 15 21 231 315
3 5 7 9 11 15 33 45 385

(Of course that he also found the minimal solution with 9 members and 231 the largest one)

Jud thinks that there are only these 5 solutions with 9 membres and that there is not any smaller one.

***

Jean-Charles Meyrignac sent the following lines the 21/5/2001:

Digging in my books to search for some documentation, I found Martin Gardner's book of Mathematical Diversions. The problem was to decompose 1 as the sum of odd inverses. This problem has been solved by S. Yamashita in 1976 (and then Jean-Charles send the 5 solutions discovered between Jud, Jaime and myself 24 years later, partially due to the lack of this info in the R.K. Guy's book).

So Yamashita is the first solver of this question 1.

***

On Octobr 5, 2020, Tatsuru Watabane wrote:

I found answers for 3) and 4).

Of course, my solution for puzzle 899, for example, {6 10 14 15 21 22 26 33 34 35 38 39 46 51 55 57 58 62 65 69 74 82 85 86 87 91 93 95 106 111 123 133 145 155 159 185 203 215 253 265 287 319 493 583 731 851 1073} is also solution of 3) and 4). But I improved Barbeau’s solution and Johnson’s solution from a different perspective. Specifically, I found the lower limit of the maximum of the denominators.

The solution of 3) is:

{6 10 14 15 21 22 26 33 34 35 38 39 46 51 55 57 58 62 65 69 77 82 85 95 111 115 118 119 123 133 141 143 145 148 155 185 188 217 221 235 259 287 295 299 319 391 403 407 413}

49 distinct positive integers no one dividing each other are used in this example. This is unique solution where all denominators are 413 or less.

The solutions of 4):

The lower limit of the maximum of the denominators is 589, and there are three solutions where all denominators are 589 or less.

{6 10 14 15 21 22 26 33 34 35 38 39 46 51 55 57 58 62 65 69 77 82 85 91 93 95 106 119 123 129 143 145 159 161 187 203 213 217 247 265 287 299 341 355 403 473 493 497 559 583 589}(using 51 numbers)

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