THE MONKEY COCONUT PROBLEM REDUX (PART II)

A NEW MATH SEQUENCE FROM THE DIOPHANTINE EQUATION 1024(x) = 15625(y) + 11529

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:

Gardner did not solve it by the known extended Euclidian algorithm finding it too difficult so instead used a simple non Euclidian method. The Euclidean method, however, was used to produce the values of x and y in Part I.

Table of Values of x and y

The following equations were obtained for x₀ and y₀:

Plugging in the values 0-16 for n affords Table I, where the yellow cells correspond to values based on y = 2^k − 1, where k ≥ 10:

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

Construction of the Sequence starting with the Odd Number 1

The numbers in yellow (x values), however, correspond to a new sequence which may be generated via the route shown in Table II. Using the initial value of y = 2⁹ − 1 we can convert the corresponding x, a non-integer value, into the next higher x by doubling x and adding 4. The result is that column 4 is equivalent to column 2 on line 2. Thus doubling and adding 4 to each x results in generation of the integer sequence:

where k is the superscript of the y = 2^k − 1 equation.

In addition, the sequence may be generated by adding 2ⁿ(5⁶) to each subsequent value of x_k starting at x₁₀ = 15621. (Note that 5⁶ is shorthand for 15625 while the arrow denotes continuation of the sequence on the next line):

An interesting property of this sequence is that at exactly x₁₆ when x₁₆ contains 5 nines each subsequent number takes on the value [(2^m − 1)×10⁶ + 999996] with m taking on all values ≥ 0. Thus, 2^m − 1 are the leftmost digits of each x_k. For example, the leftmost digits of the two next subsequent values x₂₅ and x₂₆ are 511 and 1023, respectively followed by 999996.

Finally, each term in the sequence can be defined by the following equation:

where j ≥ 1. At j = 1 the rightmost term becomes 0 giving x₁ = 15621, the first number in the sequence. In addition, a quick check shows that when j = 5, x₅ = 15621 + (15625 × 15) = 249996.

Thus, it appears that the monkey coconut problem, in actuality the Diophantine equation 1024x = 15625y + 11529, still has many surprises in store. One of them is that a general solution to this sequence has been found which can generate an infinite number of Diophantine based sequences depending on the value of odd n (see Part IV). Infinite odd n's means infinite sequences.

Go to Part III - A second new sequence.
Go back to Extended Euclidean algorithm Part I for solving the Diophantine equation.
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