THE MONKEY COCONUT PROBLEM REDUX (PART III)

A SECOND 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:

1024x = 15625y + 11529

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. In addition, a separate web page produced a new sequence Part II, which though similar to the one being constructed on this page, varies by a factor of 3.

Table of Values of x and y

The following equations were obtained for x₀ and y₀:

x₀ = 1024(−4 + 15625n)
y₀ = 15625(1 − 1024n) = -15625(−1 + 1024n)

Plugging in the values 0-16 for n affords Table I, where the yellow cells correspond to values based on y = 3×2^k − 1, where k ≥ 10:

Table I
n	x	y
0	−4	1
1	15621	1023
2	31246	2047
3	46871	3071
4	62496	4095
5	78121	5119
6	93746	6143
7	109371	7167
8	124996	8191
9	140621	9215
10	156246	10239
11	171871	11263
12	187496	12287
13	203121	13311
14	218746	143353
15	234371	15359
16	249996	16383

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

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 =3×2¹⁰ − 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 line 1 is equivalent to column 2 line 2. Thus doubling and adding 4 to each x results in generation of the integer sequence:

x₁₀, x₁₁, x₁₂, x₁₃, x₁₄, x₁₅, x₁₆, x₁₇x₁₈, x₁₉, ...x_k

where k is the superscript of the y = 3×2^k − 1 equation.

Table II
y	x	2x	2x+4
3×2¹⁰ − 1	46871	93742	93746
3×2¹¹ − 1	93746	187492	187496
3×2¹² − 1	187496	374992	374996
3×2¹³ − 1	374996	749992	949996
3×2¹⁴ − 1	749996	1499992	1499996
3×2¹⁵ − 1	1499996	2999992	2999996
3×2¹⁶ − 1	2999996	5999992	5999996
3×2¹⁷ − 1	5999996	11999992	11999996
3×2¹⁸ − 1	11999996	23999992	23999996
3×2¹⁹ − 1	23999996	47999992	47999996

Construction of The Sequence Starting with the Odd Number 3

In addition, the sequence may be generated by adding 2ⁿ(3)(5⁶) to each subsequent value of x_k starting at x₁₀ = 46871. (Note that 5⁶ is shorthand for 15625 while the arrow denotes continuation of the sequence on the next line):

	2⁰(3)(5⁶)		2¹(3)(5⁶)		2²(3) (5⁶)		2³(3)(5⁶)		2⁴(3)(5⁶)		2⁵(3)(5⁶)		2⁶(3)(5⁶)		2⁷(3)(5⁶)
46871		93746		187496		374996		749996		1499996		2999996		5999996

→

		2⁸(3)(5⁶)		2⁹(3)(5⁶)		2¹⁰(3)(5⁶)		2¹¹(3)(5⁶)		2¹²(3)(5⁶)		2¹³(3)(5⁶)
	11999996		23999996		47999996		95999996		191999996		383999996		767999996	....

An interesting property of this sequence is that at exactly x₁₆ when x₁₆ contains 5 nines each subsequent number takes on the value [(3×2^{m − 1} − 1)×10⁶ + 999996] with m taking on all values ≥ 1. Thus, 3×2^{m − 1} − 1 are the leftmost digits of each x_k. For example, the leftmost digits of the two next subsequent values x₂₅ and x₂₆ are 1535 and 3071, respectively followed by 999996.

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

x_j = 46871 + [3×5⁶ × (2^{(j − 1)} − 1)]

where j ≥ 1. At j = 1 the rightmost term becomes 0 giving x₁ = 46871, the first number in the sequence. In addition, a quick check shows that when j = 5, x₅ = 46871 + (3×15625 × 15) = 749996.

Although two sequences have been constructed from the Diophantine equation 1024x = 15625y + 11529, there is a general solution to the production of the Diophantine based sequences Part IV which can generate an infinite number of sequences.

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