9 1 1

[Scott]

6 6 6 + 2 1 6 + 3 1 1 + 2 0 2 0 = 3 2 1 3

Nov.7th is usually day 3 1 1 of the year, except in a leap year.

This year Nov.7th starts after 3 1 1 days of 2 0 2 0 are completed.

8 6 + 1 1 0 7 + 2 0 2 0 = 3 2 1 3

[Scott]

My wifes maiden name has a sum of 9 1 8

The first occurrence of 9 1 8 is flanked with 5 5 9,

5 5 9 9 1 8 5 5 9



1 5 9 0{my name) digits + 5 5 9 digits

after the completion of the first occurrence of 8 6, is the 9

in the second 5 5 9 that flanks 9 1 8, and that 9 is digit # 2 2 2 4,

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


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

In English,

God = 1 5 6
Jesus Christ = 9 0 6

[Scott]

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

1 9 8 4 occurs first starting right after digit # 3 3 5 7

Counting 3 3 5 7 digits after the completion of the first

occurrence of 9 8 7{p.11,post # 102), puts us at digit # 4 8 6 9,

which is the completion of the first occurrence of 1 5 9 0



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

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


6 6 6 + 2 1 6 digits stand between

the first occurrence of 1 9 9 5 and digit # 1 5 9 0


123 45 6 789 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

First occurrence's of 1 5 9 0 and 1 3 1 8{Rev.13:18)

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

Those 4 5 0 digits + 2 1 6 digits lands on digit # 5 5 3 1

Digit # 5 5 3 1 + 1 1 0 7 + 2 0 2 0 digits = digit # 8 6 5 8

8 6 5 8 is the English sum of the words in Ivan Panin's N.T. of Rev. 13:18

[Scott]

One word in Panin's translation of Rev. 1 3 : 1 8,

is old english, the word "hath", and if changed to "has",

the sum of the words are now 8 6 0 4

Count the number of the beast = 1 6 8 0

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

The first 2 3 4 6 completes its occurrence at digit # 2 6 3

The number 8 6 0 4 completes its first occurrence at digit # 8 8 6 7

Digit # 2 6 3 + 8 6 0 4 digits = digit # 8 8 6 7

3 . 1 4 . . . . . 2 3 4 6(. . . . . 8 6 0 4 digits . . . . . 8 6 0 4)

8 6 0 4 completes its first occurrence 8 6 0 4 digits

after the first occurrence of 2 3 4 6

If we make one more change and remove the word "for" from

"for it is the number of a man",

Here is the wisdom, who has understanding, let him
count the number of the beast, it is the number of a man,
and his number is six hundred sixty six

then the sum of the verse is 8 3 7 0, and this is what we see

at the first occurrence of 8 3 7 0,

2 3 6 8 3 7 0

[Scott]

Covid = 3 1 8

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

3 . 1 4 . . . . . . . . . . . . . . . . 9 1 1 3 1 8 6 6 6

"Coronavirus disease" {Covid-19)= 3 1 8 + 6 6 6 + 3 1 8 { 1 3 0 2 )

Including the digits of the first occurrence of 8 6

as the first 2 digits to reach digit # 1 3 0 2,

leaves, 3 1 8 + 9 1 1 digits,

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

Here's what makes an occurrence right before the

first occurrence of 1 3 0 2,

3 . 1 4 . . . . . . . . . . . 2 0 1 9 1 3 0 2

[Scott]

November Seventh = 1 1 2 2

First occurrence's of 2 0 2 0 and 1 1 2 2,

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


The last digit of the first occurrence of 1 1 2 2 being digit # 1 2 0 2 0

is what i noticed first, then, i counted the measurement

in digits from 2 0 2 0's occurrence to 1 1 2 2's occurrence.


1 5 9 0
+
1
+
5
+
9
+
0
+
One Thousand Five Hundred Ninety
+
One
+
Five
+
Nine
+
Zero

= [ 2 0 2 0 + 1 5 9 0 + 1 1 2 1 ]

[Scott]

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

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

I was 1 0 3 1 0 days old and entering day 1 0 3 1 1 when i considered

trying to find the date Ivan Panin spoke of, by counting 6 6 6


First occurrence's of 6 6 6 and 1 0 3 1 0,

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

1 5 9 0 + Count the number of the beast{1 6 8 0) + 6 6 6 = 3 9 3 6

3 9 3 6 makes its third occurrence right before the first occurrence of 1 0 3 1 0

[Scott]

Genesis 6 : 1 8 is verse # 1 5 6,

But with thee will i establish my covenant, and you shall come into
The (4 5 0) Ark, you, your son's, your wife, and your son's
wive's with thee.


In Pi, the first occurrence of number 4 5 0{Noah's Ark)

is immediately after digit # 1 5 6


Genesis 1 8 : 2 5 is verse # 4 5 0,

That (would) be far from thee to do in this manner, to slay the Righteous
with the Wicked, and that{as tho) the Righteous should be as the wicked, that
(would) be far from thee,(so,) shall not the Judge of all the earth
do right(as well)


Verse # 6 6 6 tells us that Abraham lived 1 7 5 years.


Ivan Panin explained why it was Haran, not Abraham, that was born 1 9 4 8

years after Adam was created, and Abraham was born 2 0 0 8 years after Adam,

and God made a covenant with him at the 2 1 0 7 year mark, which was

4 5 0 years after The Flood ended at 1 6 5 7 years.


2 1 0 7 - 8 6 = 2 0 2 1

The place values of the letters of Jesus Christ

in the Greek alphabet{1st letter a value of 1, last letter a value of 24),

add to 8 7

2 1 0 7 - 8 7 = 2 0 2 0


If we count 2 0 2 1 digits after the first occurrence of 1 5 6,

the next 4 digits are the first occurrence of 2 0 2 1


If we count 1 5 9 0 + 6 6 6 + 2 0 2 1

digits after the completion of 8 6, we land on the 0 in

the first occurrence of 2 0 2 1

[Scott]

I know now, 29 years later that the words in Ivan Panin's translation of

Rev. 1 3 : 1 8, that gives instructions to count 6 6 6, add to 8 6 5 8.

So knowing the connection between the number 8 6 5 8

and the number 6 6 6, i subtracted those two numbers

from the number of days{1 0 3 1 1) i had lived on the night i counted 6 6 6

and immediately after searched for a number having the same mathematical

features as Enoch's 9 8 7, because i knew God picked 9 8 7

for a very good reason,


1 0 3 1 1 - [ 8 6 5 8 + 6 6 6 ] = 9 8 7

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

Turn's out i discovered, that theres only 1 number greater than 9 8 7

sharing 9 8 7's features,(p.11,post # 102)

1 x 9 x 9 x 5 = 4 0 5

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

[Scott]

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

Post # 2 4 6

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

In Greek, "Jesus Christ" = 2 3 6 8