THE MONKEY COCONUT PROBLEM REDUX (PART IV)

A GENERAL SOLUTION TO THE DIOPHANTINE 1024(x) = 15625(y) + 11529 SEQUENCES

Background

Martin Gardner's essay of the Monkey and the Coconuts [The Colossal Book of Mathematics, 2001, pp3-9] tells the story of five men and a monkey shipwrecked on a desert island with coconuts for food. The coconuts are split in a certain manner with the monkey getting a certain amount. The problem is solved using six equations which are ultimately reduced to one - A Diophantine equation with 2 unknowns:

The equation, which has been solved by the extended Euclidian method, Part I, was used to produce the values of x and y in Part II. Two sequences were constructed previously in Part II and Part III based on the Diophantine equation 1024x = 15625y + 11529. It has been found that an infinite number of sequences exists for this equation. Consequently, a general solution to the Diophantine based sequences was derived employing the odd values of x from Table I as the initial elements of a sequence and the even values as their consecutive elements. Since Table I can continue indefinitely every value of x and y in the Diophantine equation is possible. An infinite number of x generates an infinite number of sequences. There are no last x and y numbers.

Table of Values of x and y

The following equations were obtained for x₀ and y₀ and the first 17 values of each shown in Table I:

where the difference between each x_n = 15625 while for each y_n = 1024.

A General Solution to the Diophantine Based Sequences

We may begin by stating that the following general equation is used in the sequence to convert an element of x into the next element of x:

The equations for each element are set up as follows where k is the coefficient of the rightmost subscripted n and 2 is to the power of logk/log2:

where x_1n (an odd number) is the first element of the sequence equal to (−4 + 15625n) and n is its position in Table I. In addition, n is any odd number ≥ 1 and j is a very large integer. In addition, x_1n, where n is an even number, can never be the first element in any sequence since all even numbered elements belong to one and only one sequence whose initial value is odd.

To make this clearer let us look at Table II containing the subscripts of x. Here we place the first 8 odd ns, the subscripts of the initial elements, into column 1, then double those numbers, store the result in column 2 and repeat the procedure up to column 8. The upshot is that an even number (kn) in columns n₂ thru n₈ appears only once in the table as part of a particular sequence and is never part of another sequence:

As an example, 320 belongs only to the sequence with the odd number 5 residing in column n₇: Translation x₃₂₀ having the value 4999996 shown in the sequence below. Consequently, all numbers in Table I, corresponding to the integer values of x in the Diophantine equation 1024x = 15625y + 11529, will appear in a sequence. You will never see a value of x in Table I, especially an infinite sized Table I, that is not in a sequence.

Construction of The Sequence Starting with the Odd Element Number 5

The previous two sequences starting with x_1n equal to 1 and 3, respectively have been covered in Part II and Part III. A third sequence starting with x₍₁₎₅ = 78121 is shown below along with 14 subsequent elements of x_kn. In addition, the subscripts of x_kn will show the exact position of each element in Table I if the table were large enough to include all the elements in the sequence.

As one can see the sequence follows the general solution using the initial x₅ = 78121 from Table I.
(Note that 5⁶ is shorthand for 15625 while the arrow denotes continuation of the sequence on the next line):

The sequence shows that each subscript is doubled after each addition and that after the 14th addition the number 1279999996 would find itself at position 81920 in Table I. However, a perculiar thing happens at position 640 where the number to the left of the group of 9s takes on the value of 2^m − 1 (m being ≥ 1). Thus, one can figure out all subsequent elements just by concatenating 2^m − 1 to the 9s. The numbers 1,3,7,15,31,63,127 on the next line clearly show this.

Let me also add that the first element in the sequence and the value therein are both odd. i.e., x_1n = x₅ = 78121. The values of every element higher than the initial, however, are all even as well as their position in Table I, as shown in the sequence above.

Sequence Starting with the Odd Element Number 25

It was previously shown that the odd sequences where n = 1 and n = 3 generated elements having 5 consecutive 9s followed by a 6. In the sequence where n = 5, however, each sequence generated elements having 6 consecutive 9s followed by a 6. With n = 25 and any of its multiples the number of 9's is 7 followed by a 6. Thus, taking the value in position 25 of Table 1, equal to (−4 + 15625×25) = 390621, we can generate 25^th sequence (up to the first 15 elements) using the above general solution:

The Number of Consecutive Nines

Thus, at some point in the sequence (element positions 7, 8 or 9) the number of consecutive 9s varies with the odd number as tabulated in Table III. This search was done up to n = 100, so it's probably a valid assumption for higher n.

In summary four sequences have been constructed from the Diophantine equation 1024x = 15625y + 11529, but it would take an infinite amount of time to construct all the rest. Too many and not enough time.

Go back to Part III.
Go back to Part II.
Go back to Extended Euclidean algorithm Part I for solving the Diophantine equation.
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