Squares of 8 at N decimals of precision.
Buckle up, this one is odd. Odd enough for me to garner a few new terms.
Chains; The squares of 8 have 4 chains that have 4 differing
math mathematical patterns
Directional addition; Slope, Vertical and Horizontal addition. Patterns within
the physical structure of change over time.
Chunks; 1 completed pattern cycle 0.012345679012345679 has 2 chunks of
012345679
Chunk Summation; product of all values within a given chunk of patterns
Summation Distance; Value of (Summ2 – Summ1) Reduced through Chunk Summation.
N Values, The value of N when a chunk is completed.
N Value Summation, Summation on N Values to 2 digits [Single pass min] ,
(ex:
1 = 1, 10 = 1, 19 = 10, 199 = 19 , 1999 = 28, 2080 = 10, 2089 = 19)
Double
digit summation was chosen as all N value summations equal 1 with enough
iterations.
You will notice there are only a few possible solutions; 1, 10, 19, 28, 37, 46. The majority of solutions will equal 10. For this reason we will use 10 as the background or 0.
Lets look at the way I divided the solution up;
Chain D
This chain is always a single digit 4
Chain C
This is the simplest of the expanding chains so we shall start with the below;
Chain A’s Chunk; 320987654N
Slope 4’s Chunk; 320987654N
Starting @ N = 2 we have the digit 4. @ N = 3, the digit 4 drops down and the next highest digit is placed to the left to form 54 @ N = 3. This process is repeated until the chunk is complete @ N = 10.
Please note that all the digits with the exception of 1 is utilized in the 2 chunks.
| 4 |
| 54 |
| 654 |
| 7654 |
| 87654 |
| 987654 |
| 0987654 |
| 20987654 |
| 320987654 |
Chain A
Lets first discuss the pattern of growth observed as N is
increased;
Please note this chunk starts @ N = 2.
Chain A’s Chunk; 790123456
N
Slope 1’s Chunk; 789123456N
Please note that Chain A’s chunk uses all digits except 8.
Please note that Slope 1’s chunk uses all digits except 0.
As you can see, Chain As chunk begins with the digit 7 @ N = 2. At N = 3, the 7
is dropped down and the next highest numeral is added to the right creating
0.78.
| N Val | Chain A (# decimal digits = N – 1) |
| N = 1 | 0. |
| N = 2 | 0.7 |
| N = 3 | 0.78 |
| N = 4 | 0.789 |
| N = 5 | 0.7901 |
| N = 6 | 0.79012 |
| N = 7 | 0.790123 |
| N = 8 | 0.7901234 |
| N = 9 | 0.79012345 |
| N = 10 | 0.790123456 |
Chain B
Chain B has 1 features;
Chunk; 678901245
| N Val | A | diff | B | diff | C | D |
| N = 1 | 6 | 4 | ||||
| N = 2 | 7 | = | 7 | -3 | 4 | 4 |
| N = 3 | .8 | = | 8 | -3 | 5. | 4 |
| N = 4 | ..9 | = | 9 | -3 | 6.. | 4 |
| N = 5 | …1 | +1 | 0 | -3 | 7… | 4 |
| N = 6 | ….2 | +1 | 1 | -3 | 8…. | 4 |
| N = 7 | …..3 | +1 | 2 | -3 | 9….. | 4 |
| N = 8 | ……4 | = | 4 | -4 | 0…… | 4 |
| N = 9 | …….5 | = | 5 | -3 | 2……. | 4 |
| N = 10 | ……..6 | = | 6 | -3 | 3…….. | 4 |
Chain B and Chain D have only 1 digit.
At N = 5 in Chain B the value of the digit reaches 10. As
this is only a 1 digit column the extra 1 is pushed left to Chain A. This
causes Chain A to skip the digit 0.
At N = 8 in Chain C the value of the left most digit reaches 10. As the extra digit would lie outside of the boundary of (N – 1) digits, the 1 is pushed left to Chain B. This causes chain B to skip the digit 3.
Lets look at Chain C at N = 51 to N = 55
| N Value | # of Digits (N – 1) | |
| 51 | 87654320987654320987654320987654320987654320987654 | 50 |
| 52 | 987654320987654320987654320987654320987654320987654 | 51 |
| 53 | 0987654320987654320987654320987654320987654320987654 | 52 |
| 54 | 20987654320987654320987654320987654320987654320987654 | 53 |
| 55 | 320987654320987654320987654320987654320987654320987654 | 54 |
| N Value | Chain Value | # of Digits | |
| 52 | 987654…. | 51 | |
| Add +1 edge # | 100000….. | ||
| 53 | 0987654…. | 52 |
As shown above, adding +1 to the 9 creates an extra digit that can not be contained within the 52 digits allotted for the next N iteration. This additional +1 is passed on to Chain B @ N = 53.
Memorizing all answers for squares of 8 all decimal precisions is as simple as these few rules:
1) Memorize the sequence of Chain B, [678901245]
2) Select a precision, Example: N = 10
3) Chain A is N – 1 digits long, Example: Chain Length = 9 digits
A) If Chain B digit = 0, 1, OR 2 then the right most digit of Chain A is the Chain B digit + 1
B) If Chain B digit = 4, 5, 6, 7, 8, OR 9 then the right most digit of Chain A is the Chain B digit
Example; @ N = 10, Chain B = 6. The right most digit of Chain A is 6
4) Chain C is N – 1 digits long, Example: Chain Length = 9 digits
A) If Chain B digit = 4 then the left most digit of Chain C is the Chain B digit – 4
B) If Chain B digit /= 4 then the left most digit of Chain C is the Chain B digit – 3
Example; @ N = 10, Chain B = 6. The left most digit of Chain C is 3
5) Memorize the sequence of chains A [790123456] and C [320987654];
A) For Chain A, start with the digit from step #3 and list N – 1 digits heading left.
B) For Chain C, start with the digit from step #4 and list N – 1 digits heading right.
6) Add the digit 4 to the end for Chain D
Note, for N < 5 Chain A = (nil), 7, 8, & 9
Quick Rules;
Chains A, B, & C are 9 digits long.
Answer Length = N * 2
Chain A & C Length = N – 1
Chain B & D Length = 1
Chain A starts with 7 and has no 8 (N > 4) [790123456]
Exceptions;
If Chain B = 8 then Chain A’s right most digit is 8
If Chain B = 9 then Chain A’s right most digit is 9
Chain B starts with 6 and has no 3 [678901245]
Chain C starts with 3 and has no 1 [320987654]
| 0.7901234567 | 7 |
| 0.79012345678 | 8 |
| 0.790123456789 | 9 |
| 0.7901234567901 | 0 |
Add Section On Summation Of N values .. etc
Above is information on the values of N = 1 => 280
The leftmost column shows the number of completed chunks @ a particular N value. For reasons unknown, chunks 0 & 11 only use 1 location, unlike the remainder which group in batches of 10.
Chunk Sums: A 2 digit summation of the values of the chunk count. Example; 265 +> 2 + 6 + 5 = 13
N Values: The number of repeating decimal digits
Sum of N Values: The sum of all N values for a given section.
Example; 10 + 19 + 28 + 37 + 46 + 55 + 64 + 73 + 82 + 91 = 505 after 1 iteration of summation of the digits 505 => 5 + 0 + 5 = 10. There appears to be “dividers” at N = 1, 10, 100, 1000, etc
Summation of N digits values: Summation of N Values to 2 digits.
Example; 280 => 2 + 8 + 0 = 10.
You may notice that all Sum of N Values can use iterative summation to reach the number 1 which is the reasoning behind stopping the summation @ 2 digits. With the exception of the first group, all Sum of N values will be separated by 900 units.
At chunk #12 another pattern begins to emerge. For the Chunk Sum block 165 the summation of N value digits all = 10. The next Chunk Sum block 265 The first value of N value digits = 19 while the remainder all = 10. This pattern emerges to be a growing count where the first Chunk Sum block = 0, the next Chunk Sum block = 1 and so on. As this pattern grows the contrasting numbers grow in the pattern ( 10 => 19 => 28 => 37 => 46 => etc where the only used numbers are the expanded summations of 10.
Below is information on the values of N = 129619 => 129880
The final number shown below would be 129,879 digits long!
Squares of 4
Squares of 4 have 3 chains;
Chain A Chunk: 197530864
Chain B Chunk: 135802469
Chain C Chunk: 136
Slope A: 197420863
Slope B: 964208531
Slope C: 136
(Squares of 2) * 4 =
(Squares of 4)
(Squares of 2) * 16 = (Squares of 8)
(Squares of 4) * 4 = (Squares of 8)
Squares of 5
Squares of 5 have 3 chains;
Chain A Chunk: 308641975
Chain B Chunk: 246913580
Chain C Chunk: 25
Slope A: 208531964
Slope B: 246913580
Slope C: 25
Squares of 7
Squares of 7 have 3 chains;
Chain A Chunk: 604938271
Chain B Chunk: 283950617
Chain C Chunk: 29
Slope A: 493827150
Slope B: 716059382
Slope C: 29
Cubes of 3;
3s to the 4th
3s to the 5th
3 chains;
Chain A Chunk: 020576131687 & 2427983539094650
Chain B Chunk: 246913580
Chain C Chunk: 020576131687 & 2427983539094650
Slope A: 301522633744855966077189290
Slope A (Bold): 0 1 22 33 44 55 66 77 89 90
Slope A (Reg ): 3 5 6 7 8 9 0 1 2
