Part 2: The Quran's Number 19
Date: 09/02/04

Part 2: The Quran's Number 19

Part 1
Part 3

Analysis of the mathematic formulae necessary to find multiples of 19.

Let's begin with a simple premise that:

Number 19, the number upon which all of the Quran is based, represents two octaves plus the fifth of the third octave of 7 notes.

If the premise is adhered to, each third octave will have only 5 notes while the first two will have 7 notes.

As each three octaves progress to the next three
octaves, a certain displacement of notes occurred in this manner:

Octave 19x1 required moving [0] notes to 2nd octave.
Octave 19x2 required moving [0] notes to 3rd octave.
Octave 19x3 required moving [2] notes to 4th octave.
Octave 19x4 required moving 2+2=[4] notes to 5th octave.
Octave 19x5 required moving 2+2+2=[6] notes to 6th octave.
Octave 19x6 required moving 2+2+2+2=[8] notes to 7th octave.

Rule:

Displacement advances by 2 in each octave
which means the displacement number will seem
to move backwards or counterclockwise, through
the octaves.

19x1 =19
1st octave = 1,2,3,4,5,6,7
2nd octave = 8,9,10,11,12,13,14
3rd octave-fifth = 15,16,[17],18,(19) .........   =1+9=1

If we are in octave 19x1 = 19 and want to know
what number octave 19x2=38 begins with, multiply 19x2 = 38 - 18 = 20

19x2 = 38
4th octave = 20,21,22,23,24,25,26,
5th octave = 34,35,27,28,29,30,31,
6th octave-fifth = 32,33,36,37,[(38)] .........   =3+8=11=2

If we are in octave 19x2 = 38 and want to know
what number octave 19x3 = 57 begins with, multiply 19x3 = 57 - 18 = 39

19x3 = 57
7th octave = 39,[40],41,42,43,44,45,
8th octave = 46,47,48,49,50,51,52,
9th octave-fifth = 53,54,55,56,(57) .........   =5+7=12=3

If we are in octave 19x3 = 57 and want to know
what number octave 19x4 = 76 begins with, multiply 19x4 = 76 - 18 = 58

19x4 = 76
10th octave = 58,59,60,[61],62,63,64,
11th octave = 65,66,67,68,69,70,71,
12th octave-fifth = 72,73,74,75,(76) .........   =7+6=13=4

If we are in octave 19x4 = 76 and want to know
what number octave 19x5 = 95 begins with, multiply 19x5 = 95 - 18 = 77

19x5=95
13th octave = 77,78,79,80,81,[82],83,
14th octave = 84,85,86,87,88,89,90,
15th octave-fifth = 91,92,93,94,(95) .........   =9+5=14=5

The rule seems to be:

1. The last number in any octave set of 3 is 19 x
the set number: Example 19x3, 19x4.

2. The first number in any octave set of 3 is
the last number minus 18.

What is the beginning number of the fifth octave 19x5=95:

1. Multliply 19x5 = last note of the octave set = 95
2. Subtract 95 - 18 = 77 = first number of the set.

What is the beginning note of thirteenth octave
19x13 = xxx if I am in octave 5:

1. Multliply 19x13 = last note of the octave set = 247
2. Subtract 247 - 18 = 229 = first number of the set.
Let's go to the thirteenth set of
octaves and see if the calculations are correct:

19x6=114
16th octave = 96,97,98,99,100,101,102,
17th octave = [103],104,105,106,107,108,109,
18th octave-fifth = 110,111,112,113,(114) .........   =1+1+4=6

19x7=133
19th octave = 115,116,117,118,119,120,121,
20th octave = 122,123,[124],125,126,127,128,
21th octave-fifth = 129,130,131,132,(133) .........   =1+3+3=7

19x8=152
22th octave = 134,135,136,137,138,139,140,
23th octave = 141,142,143,144,[145],146,147,
24th octave-fifth = 148,149,150,151,(152) .........   =1+5+2=8

19x9=171
25th octave = 153,154,155,156,157,158,159,
26th octave = 160,161,162,163,164,165,[166],
27th octave-fifth = 167,168,169,170,(171) .........   =1+7+1=9

19x10=190
28th octave = 172,173,174,175,176,177,178,
29th octave = 179,180,181,182,183,184,185,
30th octave-fifth = 186,[187],188,189,(190) .........   =1+9+0=10=1

19x11=209
31th octave = 191,192,193,194,195,196,197,
32th octave = 198,199,200,201,202,203,204,
33th octave-fifth = 205,206,207,[208],(209) .........   =2+0+9=11=2

19x12=228
34th octave = [210],211,212,213,214,215,216,
35th octave = 217,218,219,220,221,222,223,
36th octave-fifth = 224,225,226,227,(228) .........   =2+2+8=12=3

19x13=247
37th octave = 229,230,[231],232,233,234,235,
38th octave = 236,237,238,239,240,241,242,
39th octave-fifth = 243,244,245,246,(247) .........   =2+4+7=13=4

Rules:

1. What is the first number of octave 19x4:

Calculate the last number in any octave set of 3: 19 x the set number:
Example 19x4=76. The first number in any octave set of 3 is
the last number minus 18: Subtract 76 - 18 = 58

2. What numbers are the 3 octaves of 19x9?

Each set is 2 octaves of 7 notes and 1 octave of 5 notes.
Multliply 3 octaves per set x the multiplyer (9) = 27
If set 19x1=1st, 2nd and 3rd octave, then set 19x9= 3x9= octave 27.
The other 2 will be 2 less = octaves 26 and 25.

3. Finding Displacement:

Displacement advances by 2 in each octave.
It is [17] in the 3rd octave and [38] in the 6th octave.
Whatever the multiplyer is (x1,x2,x3,x4), double it:
Example: 19x1=2, 19x2=4, 19x3=6, 19x4=8...

Octave 19x1 required moving [0] notes to 2nd octave.
Octave 19x2 required moving [0] notes to 3rd octave.
Octave 19x3 required moving 2=[2] notes to 4th octave.
Octave 19x4 required moving 2+2=[4] notes to 5th octave.

Let's use these rules to calculate everything about
set 19x6217. This multliplyer (6217) is a valid
Quran multiple of 19.

Rules:

1. What is the first number of octave 19x6217:

Calculate the last number in any octave set of 3: 19 x the set number:
Example 19x6217=118123. The first number in any octave set of 3 is
the last number minus 18: Subtract 118123 - 18 = 118105

Let's construct the octave set:

19x6217=118123
xxth octave = 118105,118106,118107,118108,118109,118110,118111,
xxth octave = 118112,118113,118114,118115,118116,118117,118118,
xxth octave-fifth = 118119,118120,118121,118122,(118123) .........   1+1+8+1+2+3=1+6=7

We now know the end and beginning numbers of
this octave set of 3 but we don't know what 3
octaves are associated with these numbers.

Let's use rule 2:

2. What numbers are the 3 octaves of 19x6217?

Each set is 2 octaves of 7 notes and 1 octave of 5 notes.
Multliply 3 octaves per set x the multiplyer (6217) =18651
If set 19x1=1st, 2nd and 3rd octave, then set 19x6217= 3x6217= octave 18651.
The other 2 will be 2 less = octaves 18650 and 18649.

Let's construct the octave set again, plus the
octave numbers:

19x6217=118123
18,649th octave = 118105,118106,118107,118108,118109,118110,118111,
18,650th octave = 118112,118113,118114,118115,118116,118117,118118,
18,651st octave-fifth = 118119,118120,118121,118122,(118123) .........   1+1+8+1+2+3=1+6=7

How has the displacement of normal 7 note octaves
been displaced in these octaves?

3. Finding Displacement [xxx]:

Displacement advances by 2 in each octave set of three.
It is [17] in the 3rd octave and [38] in the 6th octave.
Whatever the multiplyer is (x1,x2,x3,x4), double it:
Example: 19x1=0, 19x2=0, 19x3=2, 19x4=4...

Octave 19x1 required moving [0] notes to 2nd octave.
Octave 19x2 required moving [0] notes to 3rd octave.
Octave 19x3 required moving 2=[2] notes to 4th octave.
Octave 19x4 required moving 2+2=[4] notes to 5th octave.

Let's construct the octave set again, plus the
octave numbers, plus displacement [xxx]:

Under Construction.

19x6217=118123
18,649th octave = 118105,118106,118107,118108,118109,118110,118111,
18,650th octave = 118112,118113,118114,118115,118116,118117,118118,
18,651st octave-fifth = 118119,118120,118121,118122,(118123),.........   1+1+8+1+2+3=1+6=7

Part 1
Part 3

Impossible Correspondence Index

© Copyright. Robert Grace. 2004