A CS has three possible fates;
a. Zero
b. Endless loop
c. Infinity.
This paper addresses the third fate and postulates that no CS can expand to infinity.
Consider the following CS, C_n is the nth term in the CS. O_o is the o_th odd term.
n O_o C_n Think of the CS as a sequence of odd-numbers separated by one
1 1 89 or more even-numbers. Let "e" be the number of even-numbers
2 268 between O_o and O_(o+1). Then.
3 134
4 2 67 (O_(o+1)) = [1.5/(2^(e-1)] * O_o [approximately}
5 202
6 3 101 For O_4 and O_5, e = 1. Then;
7 304
8 152 O_5 = (O-4) * [1.5/(2^(e-1)] = 19 * 1.5 = 28.5
9 76 = 19 * 1.5 = 28.5
10 38
11 4 19 For O_3 and O_4, e = 4. Then;
12 58 O_4 = (O-3) * [1.5/(2^(e-1)]
13 5 29 = 101 * 1.5/8 = 18.935
14 88
15 44 Obviously,
16 22 (O_10)/(O_1) = (O_2)/(O_1)*(O_3)/(O_2) * ... * (O_10)/(O_9)
17 6 11
18 34 Note that there is a fixed ratio for each "e". That is.
19 7 17
20 52 e R_e = O_o+1/O_o
21 26 1 1.5
22 8 13 2 0.75
23 40 3 0.375
24 20 4 0.1875
25 10 5 0.09375
26 9 5 6 0.046875
27 16
28 8 In the sample sequence, there are
29 4
30 2 e c_e = number of intervals with 'e' even numbers.
31 10 1 1 3 Here there are 2 intervals with 2, 3 & 4
2 2 even numbers and 3 intervals with one even
3 2 number.
4 2
Now (O_10)/(O_1) = (R_1)^(c_1)* (R_2)^(c_2) * (R_3)^(c_3) * (R_4)^(c_4)
(1.5^3) * (0.75^2) * (0.375^2) * (0.1875^2)
= 0.00939
The actual ratio is 0.1124. For the purposes of discussion, this is sufficient
correlation.
For a signifcantly large number of odd-numbere, say N, the expected distribution
of e's is;
e c_e
1 N/2
2 N/4
3 N/8
| |
n N/(2^n)
Hypothesis: For a sufficiently large N, the actual distribution will approximate
the expected distribution.
For a given M = N/2, the effect of an expected distribution is
(O_p)/O_1) = PROD {[(1.5/(2^n)]^[(M/(2^n)]}; for n = 0,1,2,3, ... , i
Assume that there is a "seed" for which the first 1,000,000,000 e's are all 1's
1.5^(1,000,000,000) is approximately 10^176,091,259. This means that,
C_1,000,000,0001 = C_1* 10^176,091,259
But if the next say 1,500,000,000 e's are distributed as above then,
C_2,500,000,001 = C_1 * 10^(-11,316,752)
HYPOTHESIS: Any seed, no matter how large, will eventually encounter a
normal sequence; ie, a sequence defined by E_n = N/(2^n); that reduces
the original seed to 1 (unless a cycle is encountered)