## The Fibonacci sequence has a pattern that repeats every 24 numbers.

Numeric reduction is a technique used in analysis of numbers in which all the digits of a number are added together until only one digit remains. As an example, the numeric reduction of 256 is 4 because 2+5+6=13 and 1+3=4.

Applying numeric reduction to the Fibonacci series produces an infinite series of 24 repeating digits:

1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 9

If you take the first 12 digits and add them to the second twelve digits and apply numeric reduction to the result, you find that they all have a value of 9.

1st 12 numbers | 1 | 1 | 2 | 3 | 5 | 8 | 4 | 3 | 7 | 1 | 8 | 9 |

2nd 12 numbers | 8 | 8 | 7 | 6 | 4 | 1 | 5 | 6 | 2 | 8 | 1 | 9 |

Numeric reduction – Add rows 1 and 2 | 9 | 9 | 9 | 9 | 9 | 9 | 9 | 9 | 9 | 9 | 9 | 18 |

Final numeric reduction – Add digits of result | 9 | 9 | 9 | 9 | 9 | 9 | 9 | 9 | 9 | 9 | 9 | 9 |

This pattern was contributed both by Joseph Turbeville and then again by a mathematician by the name of Jain.

We would expect a pattern to exist in the Fibonacci series since each number in the series encodes the sum of the previous two. What’s not quite so obvious is why this pattern should repeat every 24 numbers or why the first and last half of the series should all add to 9.

For those of you from the “Show Me” state, this pattern of 24 digits is demonstrated in the numeric reduction of the first 73 numbers of the Fibonacci series, as shown below:

Fibonacci Number | Numeric reduction by adding digits | ||

1st Level | 2nd Level | Final Level | |

Example: 2,584 | 2+5+8+4=19 | 1+9=10 | 1+0=1 |

0 | 0 | 0 | 0 |

1 | 1 | 1 | 1 |

1 | 1 | 1 | 1 |

2 | 2 | 2 | 2 |

3 | 3 | 3 | 3 |

5 | 5 | 5 | 5 |

8 | 8 | 8 | 8 |

13 | 4 | 4 | 4 |

21 | 3 | 3 | 3 |

34 | 7 | 7 | 7 |

55 | 10 | 1 | 1 |

89 | 17 | 8 | 8 |

144 | 9 | 9 | 9 |

233 | 8 | 8 | 8 |

377 | 17 | 8 | 8 |

610 | 7 | 7 | 7 |

987 | 24 | 6 | 6 |

1,597 | 22 | 4 | 4 |

2,584 | 19 | 10 | 1 |

4,181 | 14 | 5 | 5 |

6,765 | 24 | 6 | 6 |

10,946 | 20 | 2 | 2 |

17,711 | 17 | 8 | 8 |

28,657 | 28 | 10 | 1 |

46,368 | 27 | 9 | 9 |

75,025 | 19 | 10 | 1 |

121,393 | 19 | 10 | 1 |

196,418 | 29 | 11 | 2 |

317,811 | 21 | 3 | 3 |

514,229 | 23 | 5 | 5 |

832,040 | 17 | 8 | 8 |

1,346,269 | 31 | 4 | 4 |

2,178,309 | 30 | 3 | 3 |

3,524,578 | 34 | 7 | 7 |

5,702,887 | 37 | 10 | 1 |

9,227,465 | 35 | 8 | 8 |

14,930,352 | 27 | 9 | 9 |

24,157,817 | 35 | 8 | 8 |

39,088,169 | 44 | 8 | 8 |

63,245,986 | 43 | 7 | 7 |

102,334,155 | 24 | 6 | 6 |

165,580,141 | 31 | 4 | 4 |

267,914,296 | 46 | 10 | 1 |

433,494,437 | 41 | 5 | 5 |

701,408,733 | 33 | 6 | 6 |

1,134,903,170 | 29 | 11 | 2 |

1,836,311,903 | 35 | 8 | 8 |

2,971,215,073 | 37 | 10 | 1 |

4,807,526,976 | 54 | 9 | 9 |

7,778,742,049 | 55 | 10 | 1 |

12,586,269,025 | 46 | 10 | 1 |

20,365,011,074 | 29 | 11 | 2 |

32,951,280,099 | 48 | 12 | 3 |

53,316,291,173 | 41 | 5 | 5 |

86,267,571,272 | 53 | 8 | 8 |

139,583,862,445 | 58 | 13 | 4 |

225,851,433,717 | 48 | 12 | 3 |

365,435,296,162 | 52 | 7 | 7 |

591,286,729,879 | 73 | 10 | 1 |

956,722,026,041 | 44 | 8 | 8 |

1,548,008,755,920 | 54 | 9 | 9 |

2,504,730,781,961 | 53 | 8 | 8 |

4,052,739,537,881 | 62 | 8 | 8 |

6,557,470,319,842 | 61 | 7 | 7 |

10,610,209,857,723 | 51 | 6 | 6 |

17,167,680,177,565 | 67 | 13 | 4 |

27,777,890,035,288 | 73 | 10 | 1 |

44,945,570,212,853 | 59 | 14 | 5 |

72,723,460,248,141 | 51 | 6 | 6 |

117,669,030,460,994 | 65 | 11 | 2 |

190,392,490,709,135 | 62 | 8 | 8 |

308,061,521,170,129 | 46 | 10 | 1 |

498,454,011,879,264 | 72 | 9 | 9 |

Thanks to Joseph Turbeville for sending “A Glimmer of Light from the Eye of a Giant” and to Helga Hertsig for bringing Jain’s discovery of this pattern to my attention.

tom barnett says

Hi,

regarding this Repeating pattern in the Fibonacci Series.

I have taken analysis of this a few stages further if you would care to take a look.

The ‘adding up to 9’ thing is just one possible pattern to discover. But there are many more to be found.

You can download some of my analysis here:

http://vbm369.ning.com/forum/topics/fibonexus-and-lucanexus-continued

scroll down and download the files. You might find the

“fib divided into fib mod9 DATA.pdf, 377 KB” file the best way in.

Kind regards,

Tom

Rex says

I was looking at the Fn sequence and noticed that even when the digits are flipped they still sum to the Fn sequence .By reversing the digits in a mirror image and subtracting there is a “new “number that is the mirror of a Fn,

A few of the sums had more calculations than a mirror flip and subtraction.

For instance 610 flipped is 1006.The digits are intact and “mirrored” like a circle but the sum isn’t flipped.

The other two involve a negative number which resolves by breaking the number up and subtracting.btw…109 also has a unique Fibonacci connection with 89.

With the Fn single digits I paired adjacent numbers to form new numbers and they still formed a Fn sequence.

987,610___789-16 =773 *377*

610.0,377_1006-773 =*233*

377,233___773-332 =441 *144*

233,144___332-441 =-109 *901*_ 90-1=*89*

144,89 ___441-98 =343 343_ 343-*233*=110_110/2=*55*

89,55_____98-55 =43 *34*

55,34_____55-43 =12 *21*

34,21_____43-12 =31 *13*

21,13_____12-31 =-19 91 9-1=*8*

13,8______31-8 =23 *3+2* =5

8,5,3,2____85-32 =53 *3+5* =8

5,3,2,1____53-21 =32 *2+3* =5

3 2,1,1____32-11 =21 *1+2* =3

2,1,1,0____21-10 =11 *1+1* =2

1,1,0_____11-10 =1 *1+0 =1

1,0 ______1-0 =1 1+0 =1

Dustin Jones says

Like the way you think but you cannot use 610 one way then change it to fit the pattern. That ruins your math equations. You can’t use it as 016 and then use it for 1006 bc that is not the inverse amd also once you get a negative number then your sequence fails when the rest are positive and you cannot change the sign bc you flip the numbers. Also as you are ascending your sequence falls out of alignment with the original at the 7th digit. It should be 4 not 5 then so on your numbers do not match from there. But that is a unique way of thinking and I feel as though you are very close to a breakthrough. I would not have seen the pattern like this.

Hello,

Very interesting indeed. I was trying to apply the Fibonacci series in a musical composition and observed the pattern. The conversion from number to musical note was just subtracting octaves (12) until the number is in [1,13] interval. Do you think this reduction is correct, for musical purposes?

Regards

The concept make sense, but you might want to give it a bit more range for a more pleasing, realistic musical interpretation. It simply reduces every note to sit within a single octave while most songs have a range of low to high that covers more than an octave. An alternate approach would be to extend your range of allowable notes to cover two octaves, with a rule that would not allow any note to be more than an octave than the one before it.

That is a good idea! Thank you very much.

Regards

Thank you very much,for repeating the pattern of Fibonacci is that a Good Ideas.

thanks……….

Despues de descubrirlo por mi propia cuenta, me puse a buscar sin encontrar nada, 2 años despues veo q si se descubrio hace muchisimo tiempo.

After discovering it on my own, I started looking without finding anything, 2 years later I see that it was discovered a long time ago.

This pattern is a 12 pattern and it’s opposite…

The polyhedron with 12 faces is a DODECAHEDRON

Plato said in his writings that the DODECAHEDRON is the shape of the entire cosmos

N.A.S.A discovered in 2003 that the shape of the Universe is a DODECAHEDRON

Man made time as we know it was “ACCIDENTALLY” created being 2 cycles of 12 representing hours and minor cycles of 5×12 = 60 for minutes and seconds…

Every face of the DODECAHEDRON(12) is a PENTAGON(5) 🙂

The secret lays in the number 9. When visiting a circle the numbers always add to 9.

360=3+6+0=9. 180=1+8+0=9. 90=9+0=9. 45=4+5=9. 22.5=2+2+5=9. 11.25=1+1+2+5=9 and so on.

Same goes for all other shapes you add up the sum of their angles and they all reduce to 9. Triangle angles 60+60+60=180=1+8+0=9

Square is 90×4=360=3+6+0=9

Pentagon 108×5=540=5+4+0=9

And so on.

Nikola Tesla was right when he said 3, 6, 9 were the key to the universe. Think in terms of energy, frequency, and vibration. Also he found that earth beats at 528Hz the sum of this is 528=5+2+8=6 6 is a very special never if squared it equals the sum of all digits excluding nine but equaling it. 1+2+3+4+5+6+7+8=36=3+6=9. 6×6=36=3+6=9

If you double any number then add the original back the sum will always equal 3 or a denomination of with the exception of 5 but that’s just a mystery waiting to be solved.

1+1=2. 2+1=3. 2+2=4+2=6 6/2=3. 4+4=8+4=12=1+2=3. But five does have its own pattern when multiplying then summing, its every 9. Weird everything seems to be connected through 9. Every third number in this sequence has a sum of 3 and every six numbers has a sum of 6 and every nine numbers has a sum of 9. Any three, six, or 9 numbers in a row will give you a sum of 3, 6, or 9. Fascinating for sure

I was with you up until that business with 5 while it’s true the answer has three reasons one it is every third number which is already divisible by 3 and from there it falls back to your original pattern mention, not that it doesn’t make for good conversation, but it’s like number games that only work because they are unique to a particular base. I do love the Fibonacci business though and the pattern of 12 and its complement. it’s as much fun as expressing Phi and 5^.5+.5 but it is again a property of base 10 being halved by 5. More interesting might be phi expressed in base 2. 🙂

Just having a bit of fun here, hoping no one means to suggest that number patterns resulting from coincidences are an indication that the universe was created by the great Mathematica. 🙂

thank you so much! this is absolutely fascinating! there are 9 major triangles in the sri yantra 🙂

My father once ran a number sequence by me that I’ve drifted back to over the past 30 years:

Start with any two, none zero, digits and note them

Sum them (if the value is greater sum those two digits ie 13 becomes 4) not result

Now sum last two digits

Continue until you get your starting numbers

This produces 5 Sequences:

11235843718988764156281911

13472922461786527977538213

14595516742685494483257314

3369663933

9999

Obviously, this is simply a numeric doodle based on reduction and the first sequence is simply reduction Fibonacci but I’ve always felt there was something there.

I’ve stretched this out to 3, 4 and 5 digit reductions (thanks to Excel) and the results show lots of patterns where factors of nine, obviously, run throughout.

Scaling up the listed article’s proposals for these other sequences is head spinning but strangely beautiful

I only really reply here as this is the first time I’ve seen my father’s time-passing exercise listed anywhere.

I discovered the same thing and also triple figures

1113598 ………….. 111

Deceived the 21 series

Plus five binary strings equals the number of letters of the alphabet

Please e-mail me back..

Fascinating. Here’s another aspect of the Fibonacci series which reveals another repetitive pattern.

I just recently discovered that if you take the Fibonacci squares and translate each number to a pitch in the one octave 7-note scale you end up with palindromatic, infinitely repeating series of pitches.

The first numbers in the Fibonnaci series of squares are 1, 1, 4, 9, 25, 64, 169

This will translate to the following pitches in one octave: 1, 1, 4, 2 (9-7), 4 (25-21), 1 (64-63), 1 (169-168).

If we continue this process, this is the pattern we end up with:

1 1 4 2 4 1 1 7 1 1 4 2 4 1 1 7 etc

Seeing the 7 as the center you end up with a 15 note palindromatic, infinitely repeating pattern. Not only that, but the pitches between the 7s make up another palindromatic, repeating pattern with the 2 as the center made of 7 pitches (1 1 4 2 4 1 1).

Numerif duction in fibonacci series is jst awesome

The true 216-digit sequence is (Naturalis Veritas, the end of the history, Massimo Nardotto, 2007):

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

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

3 3 6 9 6 6 3 9 3 3 6 9 6 6 3 9 3 3 6 9 6 6 3 9

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

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

6 6 3 9 3 3 6 9 6 6 3 9 3 3 6 9 6 6 3 9 3 3 6 9

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

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

9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9

Ref. http://en.wikipedia.org/wiki/Talk%3APi_(film)

Surprised no one mentioned that the two sets of inverse 12 digital roots create a matrix of 3-digit cells delimited by 3, 6, and 9 (terminator) every 4th place..

Not only does the content of each cell reduce to { 1, 2, 4, 8, 7, 5} in order, but the cell reductions are also inverse of the other row., totally to 9.

See below:

{1 1 2} 3 {5 8 4} 3 {7 1 8} 9

{8 8 7} 6 {4 1 5} 6 {2 8 1} 9.

cell reduction:

4 8 7

5 1 2. Recognize the pattern? Its the doubling data set {1, 2, 4, 8, 7, 5}, and it

—– —- — continues in the next 24 digits perfectly.

9 9 9

.

1st 24-digits 6 cells: {4, 8, 7, 5, 1, 2} 2nd set {4, 8, 7, 5, 1, 2} and so on, ad infinitum.

This (previous post) would suggest that the 24-digit repeating pattern is designed for:

6 sets of 3-digit data cells = 18, plus

6 delimiters (3-6-9) = 6.

18 data + 6 delimiters = 24 places.

What’s even more convincing is the STRUCTURE that the cells {data sets} and dividers {delimiters} creates and how perfectly they dovetail together with the next set of 12 places in the repeating process:

[cell} divider {cell} divider {cell} divider —fits—> [cell} divider {cell} divider {cell} divider

{cell} divider {cell} divider {cell} divider —fits—> [cell} divider {cell} divider {cell} divider

Just like the reduced cell values fit in continuation with the next set of 6 – {4, 8, 7, 5, 1, 2] –> {4, 8, 7, 5, 1, 2}.

Something dodecahedral about it.

Metatron’s Cube as a Clock and a Compass

I wanted to introduce the work of a guy that has been working with the 60 Digit Fibonacci repeat cycle and has worked it out onto Metatrons Cube as a Clock and A Compass. His work is very impressive to say the least, and I’m sure with a community like this one, and the VBM369.ning.com community many more discoveries will happen. I was able to discover the 24 digit repeat cycle from his work using Mod9. I simply used Mod9 on each Fibonacci sequence and and got a repeating sequence in just 24 digits. I see some others have found this as well using other methods.

His Name, Lucien Khan

His Book, The 216 Letter Hidden Name of God – Revealed

List of Videos on Youtube – https://www.youtube.com/user/lucienmation/videos

Summary video – https://www.youtube.com/watch?v=bNEUUfrkubM

You can find a free online document with hi-res images here – https://docs.google.com/document/d/1mVWd1aLiYZQU8VvYFBnW8kxodeYim3bYDIFfh-w42eU/pub

Link to Book on Amazon – http://www.amazon.com/216-Letter-Hidden-Name-God/dp/1492162035/

I hope you didn’t mind that I quoted your work and words in mine

http://www.symbiosissite.wordpress.com/

Thank you.

I don’t want to deny the symetrical importance of 9 (as a base), but it is rather random, like our decimal system (althought we have 10 “digits” to count with). When you add numbers you practically use 9 as a base and note the last remaining digit. Take any number (as a base) for instance 7, and you ‘ll find a similar complementary repetition: 1/7. 1/7, 2/7, 3/7/, 5/7, 8/7 (= [1]1/7), 13/7 (= [1]6/7), 21/7 (= [3[0/7) after which the remainders become 6,6,5,4,2,6,1 and of course 0. than you have a a repeating pattern of 2 x 8 numbers = 16. (You can check it: 34/7 = [4]6/7 and 55 =[7]6/7)

Nevertheless, some factors are more interesting dan others, like de Lucas numbers that generate a repeating pattern every second number: o\2 1\1 3\3 8\4 21\7 55\11 144\18 etc.

ps double the Lucas numbers reveals a repeating pattern in base 10

4, 2, 6, 8

14, 22 36 58

104, 162, 266, 428,

694, etc.

The following comment should be deleted:

March 25, 2018 at 8:07 am

ps double the Lucas numbers reveals a repeating pattern in base 10

4, 2, 6, 8

14, 22 36 58

104, 162, 266, 428,

694, etc.

thank you

ps twice the Lucas numbers show a nice repeating last digit pattern every 5th number (in base 10)

4, 2, 6, 8, 14, 22 36 58, 94, 152, 246, 398, 644, 1042, 1686, 2728, 4414,7142,11556,18698

The pattern is quite similar to that of the basic numbers:

0,1,2,3.4.5,

10,9,8,7,6,5

Instead of using 9 merely as a base one can also use it as a top swapping the 2 strings, starting with 8, 8, 7, 6, etc.

Combining the two in such a way that you always note the smallest difference from either the base or the top reduces the pattern to a repeating 11 numbers string (not counting zero’s)

0,1,1,2,3,4,1,4,3,2,1,1,o etc.

Applying the same method using counterpartner 11

0, 1, 1, 2, 3, 5, 3, 2, 1, 1, 0 etc,

I think the last digit should be a zero since it represents base nine:

0 1 1 2 3 5 8 4 3 7 1 8 0

0 8 8 7 6 4 1 5 6 2 8 1 0

This way one can also see the zigzag pattern(s) from right to left

Althought number reduction is a nice mathemagical trick, it is easier to simply divide the numbers by 9 (use it as a denominator) and note the remaining numerators. Example 13/9 = 1,4/9 (1,44444..) or 21/9 = 2,3/9 (2,3333…)

Even better to use Fibonacci-number 8 as the denominator, since every 6th number is divisible by 8 and every 12th by 9 because of that.

0/8

1/8

1/8

2,8

3/8

5/8

0/8 + 1

5/8 + 1

5/8 + 2

2/8 + 4

7/8 + 6

1/8 + 11

0/8 + 18

1/8 + 29

1/8 + 47

2/8 + 76

3/8 + 123

5/8 + 199

etc.

The 12 repeating pattern of remaing last digits (and it’s inverse):

0 1 1 2 3 5 0 5 5 2 7 1 0

(0 7 7 6 5 3 0 3 3 6 1 7 0)

The pattern can be read backwards as: 0, 1,-1, 2,-3, 5, …

Divided by 5 the repeating pattern is: 0, 1, 1, 2, 3, 0, 3, -2, 1, -1, 0

(0/5, 1/5, 1/5, 2/5, 3/5, 1 + 0/5, 1 + 3/5, 3 – 2/5, 4 + 1/5, 7 – 1/5, 11 + 0/5)

I am clearly not a math student, let alone teacher, since then I would have known about the modulus and i did not have to figure it out by myself. What a shame people with basic mathematical knowledge are left in the dark about what happens when you reduce (Fibonacci) numbers like that and why the 24- pattern (of remainders in mod 9) repeats like the 60 last digits / remainders in mod (10). Anyways, quite interesting to extend the procedure to the numbers in mod 1

I read that such a repeating pattern (in any mod) is called a Pisano Period.

The 24-pattern could be represented by a repeating string of 24 Fibonacci-numbers:

0 -1 -1 -2 -3 -5 -8 5 -3 2 -1 1 0 1 1 2 3 5 8 -5 3 -2 1 -1

Adding the string to the Fibonacci-sequence will result in numbers that are all divisible by 9 (and 18)

There is a nice cross-relation with the Lucas-sequence:

144 + 0 = 18 x 8

144 – 0 = 8 x 18

233 + 1 = 18 x 13

233 – 1 = 8 x 29

377 + 1 = 18 x 21

377 – 1 = 8 x 47

610 + 2 = 18 x 34

610 – 2 = 8 x 76

0 2 2 4 6 10 16

0 1 1 2 3 5 8

0 3 3 6 9 15 24

0 4 4 8 12 20 32

0 7 7 14 21 35 56

0 11 11 22 33 55 88

0 18 18 36 54 90 144 234 378 612 990

0 29 29 58 87 145 232

0 47 47 94 141 235 376

0 76 76 152 228 380 608

0 123 123 246 369 615 984

Another option is reduction by subtraction. Generating remainders and deficits, with 11 (times the Fibonacci sequence) as base and top

1+2=3

2-1=1

3+1=4

5+0=5

8+1=9

1+3=4; 3+1=4

2+1=3; 1+2=3

3+4=7; 4+3=7

6-5=5; -5+5=0

9-1=8; -1+8=7

Considering Fibonacci number 8 to be 11 -3, the repeating pattern of 10 digits is as follows:

0,1,1,2,3,5-3,2,-1,1,0

It might be more ‘primary’ to represent the repeating cycle of remainders and deficits in mod 11 as follows:

5, -3, 2, -1, 1, 0, 1, 1, 2, 3, 5

with the middle 0 remainers or deficits corresponding with all Fibonacci-numbers that are divisible by 11; every 11th number or should I say 10th ? That’s a bit of a defenitive problem when it comes to a cyclical pattern, since the last element of the previous cycle is the firsts of the next. Nr. 5 is covalent.

This might be easier:

13 = -1 +3 = +2

21 = -2 +1 = -1

34 = -3 +4 = 1

55 = +5 -5 = 0

89 = -8 +9 = +1

144 = +1 -4 +4 = +1

233 = +2 -3 +3 = +2

377 = +3 -7 +7 = +3

The results are the deficits and remainders in (Fibonacci-) modulus 11

I notice that I accidentally swapped the positive and negative charge when it comes to 55 as it dies not really matter. Better in general to start with substraction from the right in this alternate way of reducing numbers to a single digit in order to make it more methodological. This means that one has to read (and write) from the right to the left. So imagin the following ro be outlined on the right:

0 = 0 – 0

1 = 0 – 1

1 = 0 – 1

2 = 0 – 2

3 = 0 – 3

5 = 0 – 5

8 = 0 – 8

2 = 1 – 3

– 1 = 2 – 1

1 = 3 – 4

0 = 5 – 5

When reading the result in reverse one might notice the start of an alternating Fibonacci-sequence, or is it the end ?

A Happy End of 2019

The alternative way of reducing (Fibonacci) numbers is hardly documented, for some reason.

I believe it deserves a more prominant spot within number analysis, especially in regard to the Fibonacci sequence, as it reveals both remainders and deficits, as Fibonacci-numbers. Even more if it is extended to another radix

The other radix is (Lucas) mod 5. Switching the alternation:

1 -3 = -2

2 -1 = 1

3 -4 = -1

5 -5 = 0

8 -9 = -1

-1 +4 -4 = -1

-2 +3 -3 = -2

-3 +7 -7 = -3

-6 +1 +0 = -5

(Modulus 5 x Lucas: 10 5 15 20 35 55 90 145)

I used this method of numeric reduction in a Fibonacci like series, but adding the last three digits… (This creates a very interesting pattern of 3 groups of 13 digits, which then repeat.

I found that this works with any 3 digit number not containing any zero. And the pattern is always the same… 3 clearly related rows of 13 digits that then repeat. Here is an example….

1,1,1, 3,5,9,8,4,3,6,4,4,5

4,4,4,3,2,9,5,7,3,6, 7,7,2

7,7,7, 3,8,9,2,1,3,6, 1,1,8

1,1,1…….

The only numbers that produce a different pattern are any 3 digit number that contains any combination of only 3’s, 6’s, or 9’s. These produce there own unique pattern.. Still with 13 digits though.

9,9,9 produces only repeating 9’s.

I discovered that there are 18 unique 39 digit groups, 2 13 digit groups (numbers with only 3,6,9 in them, and of course 9,9,9.

In these groups every 3 digit number not containing a zero exists once, and only once.

Is there any formula where only one Fibonacci numbers is required to find next Fibonacci numbers..where this formula will use

The simplest approach to find the next Fibonacci number is to take the current number, multiply it by 1.618 and then round to the nearest integer.

As shown on my page at https://www.goldennumber.net/math/, you can compute any number of the Fibonacci series with the formulas shown in the section “Determining the nth number of the Fibonacci series.”