9 1 1

[Scott]

The first occurrence's of 6 6 6 and 2 4 7 2,

3 . 1 4 . . . . . . . . 6 6(6 . . . . . 1 0 5 9 0 . . . . . 2 4 7 2)

1 5 9 0 + 6 6 6 + 2 1 6 = 2 4 7 2

6 6 6, 1 0 5 9 0, post # 224


It's been 2 9 years since i began this work,

3 . 1 4 . . . . .(8 6 . . . 1 5 9 0 + 6 6 6 . . . 1 5 6)

3 . 1 4 . . . . . 8 6( . . . . . . 2 3 6 8 + 2 9 . . . . . . . .} digit # 2 4 7 2

3 . 1 4 . . . . . 8(6 . . . . . . 2 3 6 8 . . . . . . 6 6 6)

so it works out that the completion of the first occurrence of

the number 6 6 6 and digit # 2 4 7 2,

have 2 9 digits standing between them.


It will be the 2 6th Nov.7th including 1 9 9 5's

2 3 x 6 8 + 2 6 = 1 5 9 0

1 5 x 9 0 + 1 7 6 + 6 6 6 + 1 7 6 = 2 3 6 8

1 7 6 + 3 1 4 + 1 7 6 = 6 6 6

After the first 1 7 6 verses, The 4 5 0 ft. long Ark

is "Lift Up above the earth".

The last 1 7 6 verses begin with instruction's

to "Count 6 6 6"

[Scott]

The first 3 1 4 digits of Pi + 1 1 + 7 + 2 0 2 0 digits = digit # 2 3 5 2

Genesis chapter seven verse seventeen{verse 1 7 7) = 2 3 5 2

When the number 2 3 5 2 first occurs and completes at digit # 9 6 8 9,

this is what you see,

6 2 3 5 2 6 6

I counted towards the first occurrence of the number 2 3 5 2,

3 1 4 + 1 1 + 7 + 2 0 2 0 digits, using the first occurrence

of my name for the first 4,

3 . 1 4 . . . . .(1 5 9 0 . . . 3 1 4 + 1 1 + 7 + 2 0 2 0 + 1 5 9 0 + 6 6 6 + 2 1 6 . . . . 6 2 3 5 2)6 6

[Scott]

"You discovered the 6 6 6, 4 5 0, and 2 1 6 code"

= 2 7 9 6

The first occurrence of 2 7 9 6

3 .(1 4 . . . . 6 9 1 digits . . . . 2 7 9 6)

My name, 1 5 9 0, completes its first occurrence at digit # 4 8 6 9

3 .(1 4 . . . . 6 9 1 + [ 2 7 9 6 + 6 9 1 ] + 6 9 1 digits . . . . . 1 5 9 0)

Digit # 2 7 9 6 is the second digit in the first occurrence of

the number 2 7 9 9.

The last digit of this first occurrence of 2 7 9 9,

is the last digit of the first 2 7 9 9 position's of Pi,

(3 . 1 4 . . . . . . . 2 7 9 9 positions . . . . . . . . 2 7 9 9)

3 . 1 4 . . . . 8 6( . . . . . . 2 7 2 4 . . . . . . . . 2 7 9 9)

November Seventh Two Thousand Twenty = 2 7 2 4


"The 6 6 6 and 4 5 0 code"

= 1 5 9 0

"Count the number of the beast" {1 6 8 0) + 6 6 6 + 4 5 0 = 2 7 9 6

1 6 8 0 + 6 6 6 = [ 1 5 9 0 + 7 5 6 ]

Here's the first occurrence of 1 5 9 0 and 1 3 1 8,

3 . 1 4 . . . . . . . . . . . . . (1 5 9 0 . . . . . 4 5 0 digits . . . . . . . 1 3 1 8)


1 2 3 | 4 5 | 6 | 7 8 9 | 0

1 2 3 + 6 6 6 = 7 8 9

6 x 6 x 6 = 2 1 6

6 6 6 - 2 1 6 = 4 5 0

123 456 789 0

1 5 9 0 + 1 2 3 + 4 5 + 6 6 6 + 7 8 9 = 3 2 1 3

[Scott]

The following strings of numbers are the closest in the

first 200 million digits, that you can get to a full number set,

and it may or may not do this with each number in the set way

after the 200 million digit mark, but i really don't know if it does

or not, but this caught my attention,


1 2 3 4 5 7 8 9 0

1 2 3 4 5 6 8 9 0

1 2 3 4 5 6 7 8 0

The only numbers that can be omitted are,

6, 7, 9

6 7 9 + 9 1 1 = 1 5 9 0

1 3 + 6 6 6 = 6 7 9

1 3 x 6 6 6 = 8 6 5 8

8 6 5 8 is the english sum(a=6 b=12...) of Ivan Panin's N.T. translation

of Rev. 1 3 : 1 8

[Scott]

I counted the digit # where my name completes its first occurrence, 4 8 6 9

and subtracted the answer from 4 8 6 9 just as i did with 6 6 6,

4 x 8 x 6 x 9 = [ 1 5 6 + 9 0 6 + 6 6 6 ]

4 8 6 9 - [ 1 5 6 + 9 0 6 + 6 6 6 ] = 3 1 4 1 {post # 221)


and the same with my name,

1 x 5 x 9 = 4 5

1 5 9 0 - 4 5 = 1 5 4 5

1 5 4 5 occurs first right after digit # 6 8 9 4

6 6 6 + 2 1 6 + 6 8 9 4 + 6 6 6 + 2 1 6 = 8 6 5 8

1 5 9 0 + 5 3 0 4 = 6 8 9 4

5 3 0 4, post # 2 2 4


Digit # 1 5 4 4
+
[ 6 6 6 + 2 1 6 ]
+
[ 6 7 9 + 6 6 6 + 2 1 6 ]
+
[ 6 6 6 + 2 1 6 ]

= digit # 4 8 6 9


"Is November Seventh Two Thousand Twenty, the date of The Rapture"?

= [ 2 5 9 8{Revelation chapter Thirteen verse Eighteen) + 1 5 9 0 ]

3 .(1 4 . . . . 6 8 1 . . . . . 1)(0 5 0 . . . . 2 5 9 8 + 1 5 9 0 . . . . 1 5 9 0)

3 . 1 4(1 5 9 2 6 . . . . 6 7 9 . . . . . 1)0 5 0

Richard Amiel McGough = 1 0 5 0

<><><><><><><><><><><><><><><><>


1 9 6 3 + 2 0 2 0 = 3 9 8 3

The first occurrence of the number 3 9 8 3 begin's at digit # 2 0 2 0,

so 1 9 6 3 digits later, is digit # 3 9 8 3


3 .(1 4 . . . . . . . 2 0 2 0 . . . . . . . 3)(9 8 3 . . . . . . 1 9 6 3 . . . . . } digit # 3 9 8 3


The first occurrences of 8 6 and 6 6 6,

3 . 1 4 . . (86 . . . 2 3 6 8 . . . 6 6)6

The first occurrences of 6 6 6, 1 9 6 3, and 1 5 9 0,

3 . 1 4 . . . . . 6 6(6 . . . . 2 3 6 8 . . . 1 9 6 3)(. . . 5 7 . . .)1 5 9 0

1 9 6 3 + 5 7 = 2 0 2 0

[Scott]

The words in Panin's translation of Rev. 1 3 : 1 8 {count 666),

have an English sum of 8 6 5 8.

In Pi the first occurrence of 6 6 6 begins right after digit # 2 4 3 9

2 4 3 9 + 7 9 8 + 9 8 7 + 1 9 9 5 + 2 4 3 9 = 8 6 5 8

8 6 5 8 - 6 6 6 = [ 1 9 9 5 + 5 9 9 7 ]

Year 1 9 9 5 was 5 9 9 7 years from Adam's creation.

(1995, post #'s 102, 109, 115)


6 6 6 completes its first occurrence at digit # 2 4 4 2

2 4 4 2 + 6 6 6 + 2 4 4 2 + 6 6 6 + 2 4 4 2 = 8 6 5 8

The 70th anniversary of Israel, and the end of the generation that

would see signs in the Sun Moon and stars "before" it ended,

and it did, was on,

"May Fourteenth Two Thousand Eighteen" = 2 4 2 4 {page 21, post # 202),

2 3 6 8 + 1 9 9 5 years after The Flood ended.

[Scott]

7 9 8 + 9 8 7 + 1 9 9 5 = 3 7 8 0

First occurrence of 3 7 8 0,

3 . 1 4 . . . . . . . . . . . . . . 4 0 4 4 3 7 8 0

8 6 + 2 3 6 8 + 1 5 9 0 = 4 0 4 4

6 6 6 + 3 3 7 8{Saturday November Seventh Two Thousand Twenty) = 4 0 4 4

1 5 6 + 9 1 8 + 1 3 8 0 + 1 5 9 0 = 4 0 4 4

9 1 8 and 1 3 8 0 are the sums of my wife and son's name.

1 5 6 + 9 0 6 + 9 8 7 + 1 9 9 5 = 4 0 4 4

Count the number of the beast = 1 6 8 0

1 6 8 0 + 6 6 6 or 1 5 9 0 + 7 5 6 = 2 3 4 6

First occurrence of 2 3 4 6,

3 . 1 4 . . . . . 2 3 4 6( . . . . . . . 3 7 8 0 . . . . . . . . . . digit # 4 0 4 3

3 . 1 4 . . . . . 2 3 4 6( . . . . . 3 5 1 6 . . . . . . digit # 3 7 8 0

7 0 weeks of years began 3 5 1 6 years after Adam and Eve were created.

3 5 1 6, page 20 post # 1 9 5

[Scott]

I looked for the first occurrence of 6 6 6 4 5 0,

to see if 1 5 9 0 might happen to appear close by,

it didn't, but this number showed up right before,

1 9 6 3 6 6 6 4 5 0


The first time 1 5 9 0 occurs, it completes its

occurrence at digit # 4 8 6 9.


1 + 2 + 3 + 4 . . . . .+ 3 6 = 6 6 6,

First occurrence of 3 6 followed by 6 6 6,

3 .(1 4 . . . 4 8 6 9 . . . 1 5 9 0)( . . . 5 x 4 8 6 9 + 3 2 1 3 . . . 3 6)6 6 6

123 45 6 789 / 0

I knew immediately 29 yrs ago, that it was God leading

me to count 6 6 6

after seeing it structured in the core of the number set,

which created the 3 2 1 3,

and then he had me search for a number that shares the

same mathematical features with 9 8 7 {Enoch).

1 5 9 0 begins to occur first at digit # 4 8 6 6

3 .(1 4 . . . . . 6 6 6 + 3 2 1 3 + 9 8 7 . . . . . 1) 5 9 0


At the end of my 1 0 3 1 0 th day of life, i read for the first time

in Panin's Bible chronology book, that he believed

numerics clearly declare a date for The Blessed day(Rapture),

and right before he made the claim he stated that the word "the"

was supposed to precede the word "wisdom" in Rev.13:18, so i

started taking a look at the 6 6 6 climb from 123 to 789.

2 nights later i discovered that the 4, 5, 0 from the number set

is utilized by taking the result of counting 6 6 6, from 6 6 6,

and also made the 1 9 9 5 discovery.

Here's the first occurrence of 1 0 3 1 0 in Pi, and the 4 digit

number appearing right before it,

3 9 3 6 1 0 3 1 0

1 5 9 0 + 1 6 8 0 + 6 6 6 = 3 9 3 6


The first occurrence of 1 0 3 1 2 is followed by 0 3 1 1 9 9 5

Nov.7th{day 3 1 1) of 1 9 9 5 was the first date

i had my eye on as a possible Rapture date.

[Scott]

First occurrence of my name,1 5 9 0{582 + 462 + 546),

3 .(1 4 . . . . . . 4 8 7 0 . . . . . . . 7 5 1 5 9 0 6)

7 5 6, post # 204



First occurrence of 6 6 6 completing at digit # 2 4 4 2,

and 4 8 7 0 starting at digit # 5 4 6 6

3 .(1 4 . . . . 2 4 4 2 . . . 6 6 6)( . . . 1 5 9 0 + 1 4 3 4 . . . 4)8 7 0

Nineteen Sixty Three = 1 4 3 4

[Scott]

7 0 x 7 = 4 9 0{7 0 weeks of year's)

The Time of Jacob's Trouble takes place in week 7 0

The number 4 9 0 makes its first occurrence starting at digit # 9 0 7

Now compare with page 21 post #'s 202, 203


The first 3 2 1 3 digits of Pi, minus the first 7 5 because that's

where 8 6 completes its first occurrence, leaves 3 1 3 8 digits.

When i looked up the first occurrence of the number 3 1 3 8 this is what i saw,

3 2 1 3 1 3 8

If i subtract the first 8 6 digits instead of 7 5,

1 1 0 7 + 2 0 2 0 digits are left.