Who was Faboccani?
Did you mean Fibonacci ?
a 12th Century Italian mathematician, most famous for discovering the pattern and working with the numbers 1,1,2,3,5,8,13,21,...
named the "Fibonacci Numbers" after him.
This huge website can keep you busy for weeks.
http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fib.html
Leonardo Fibonacci, originally known as Leonardo of Pisa, was an Italian merchant and mathematician who contributed much to the field of algebra, Euclidian geometry, Diophantine equations, and number theory. He was instrumental in introducing the Hindu-Arabic number system to Europe. Among his many writings was the Liber Abaci, published in 1202, which contained many problems, the most famous of which, about rabbits, led to what we refer to today as Fibonacci numbers or the Fibonacci sequence. It has been quoted many ways in historical literature but basically asks, "How many pairs of rabbits can be produced from a single pair in a year, each pair producing a new pair after the second month and every month thereafter? The accumulation of rabbits looks like the following.
End of Month No.------1.....2.....3.....4.....5.....6
Pair No. 1-----------------1.....1.....1.....1.....1.....1
Pair No. 2-----------------.............1.....1.....1.....1
Pair No. 3----------------.....................1.....1.....1
Pair No. 4----------------............................1.....1
Pair No. 5---------------.............................1.....1
Pair No. 6---------------....................................1
Pair No. 7--------------.....................................1
Pair No. 8--------------.....................................1
Total........................1.....1.....2.....3.....5......8.....13.....21.....34.....55.....89.....144.....233.....377
As you can readily see, the sequence continues 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377,........n, each succeeding term being the sum of the previous two terms expressed by Fn = F(n-1) + F(n-2). The initial terms are F1 = 1 and F2 = 1.
Each successive pair of Fibonacci numbers are relatively prime, i.e., they have no common factors other than 1.
Each Fibonacci number is defined in terms of the recursive relationship, Fn = [F(n-1) + F(n-2)]. To determine the 10th, 100th, or 1000th Fibonacci number, one would normally have to compute the previous 9, 99, or 999 numbers in order to compute the one desired. It is only natural therefore, to ask whether there is a simple, or complex, expression out there someplace that would allow us to calculate any Fibonacci number desired.
Search no more and surprisingly, to me at least, it involves the equally famous Golden Ratio or Golden Number,
t = (1 + sqrt5)/2, and its reciprocal, 1/t. The expression is simply
Fn = (t)^n - (-t)^-n = (t)^n - (-1/t)^n
............sqrt(5)............ sqrt(5)
where t = the famous 1.618033988749894....... or simply 1.618, as we normally use it.
The ratios of any one Fibonacci number to the previous number progressively close in on the Golden Ratio, 1.6180, = (sqrt5 + 1)/2. Surprisingly, individual terms of the Fibonacci sequence also derive from the Binet expression
Fn = [((1 + sqrt5)/2)^n - ((1 -sqrt5)/2)^n]
............................sqrt(5)
This amazingly simple expression involving square roots and powers of an irrational number does, in fact, produce the numbers in the series.
Other ways of expressing the same thing are
Fn = [(1 + sqrt5)^n - (1 - sqrt5)^n]
.......................sqrt(5)
= [t^n - (1 - t)^n]
.........sqrt5
= [t^n - (-1/t)^n]
........sqrt5
The derivation of these may be found in any good book on number theory or recreational mathematics.
Considering the first expression, lets see what we get:
n................t^n...................(-1/t)^n..............t^n - (-1/t)^n.............[t^n - (-1/t)^n]/sqrt5
1..........1.6180339............-.6180339...............2.2360679...........................1
2..........2.6180339...........+.3819660...............2.2360679...........................1
3..........4.2360679............-.2360679...............4.4721359...........................2
4..........6.8541019...........+.1458980...............6.7082039...........................3
5........11.0901699............-.0901699.............11.1803398...........................5
6........17.9442719...........+.0557280.............17.8885438...........................8
7........29.0344418............-.0344418.............29.0688837..........................13
8........46.9787137...........+.0212862.............46.9574275..........................21
How elegant.
Closer examination of the series of numbers will lead you to some surprisingly interesting and unique relationships between the Fibonacci numbers Fn = F(n-1) + F(n-2).
As n increases, the ratio of F(n+1)/Fn approaches the Golden Ratio, (1 + sqrt5)/2 = 1.618....
The sum of the squares of two adjacent Fibonacci numbers is equal to a higher Fibonacci number according to Fn^2 + F(n+1)^2 = F(2n+1). For instance, the 4thFn^2 + the 5thFn^2 = the F(2(4) + 1) = 9th Fn or 3^2 + 5^2 = 34, the 9th Fn.
The product of two alternating Fibonacci numbers minus the square of the one in between is equal to +/- one as expressed by F(n-1)F(N+1) - Fn^2 = (-1)^n. For instance, 8x21 - 13^2 = 168 - 169 = (-1)^7 = -1.
The sum of the cubes of two adjacent Fibonacci numbers minus the cube of the preceding one is equal to a higher Fibonacci number as expressed by Fn^3 + F(n+1)^3 = F3n. For instance, F4^3 + F5^3 - F3^3 = 3^3 + 5^3 - 2^3 = 27 + 125 - 8 = 144 = F12 = 144.
The sum of the squares of a series of Fibonacci numbers starting with F1 and ending with Fn is equal to the product of Fn and F(n+1) as expressed by F1^2 + F2^2 + F3^2 + ......Fn^2 = FnF(n+1). For instance, 1^2 + 1^2 + 2^2 + 3^2 + 5^2 = 1 + 1 + 4 + 9 + 25 = 40 = 5x8.
The sum of a series of Fibonacci numbers starting with F1 and ending with Fn is equal to a higher Fibonacci number as expressed by F1 + F2 + F3 + .......Fn = F(n+2) - 1. For instance, 1 + 1 + 2 + 3 + 5 + 8 = 20 = 21 - 1.
The sum of any 10 consecutive Fibonacci numbers is 11 times the 7th term of the 10 numbers.
For instance, the sum of the 4th through 13th numbers, 3,5,8,13,21,34,55,89,144,233, is 11x55 = 605.
The sum of any number of consecutive Fibonacci numbers is given by S[Fn1-->Fn2] = F(n2+2) - F(n1+1).
For instance, the sum of the 5th through 10th numbers, 5,8,13,21,34,55, is 144 - 8 = 136.
The sum of the first n even terms of Fibonacci numbers is given by F2 + F4 + F6 + F8 + ....F2n = F(2n+1) - 1.
For instance, the sum of the first 6 even terms is therefore F(2(6) + 1) - 1 = F(13) - 1 = 233 - 1 = 232.
The sum of the first n odd terms of Fibonacci numbers is given by F1 + F3 + F5 + ....F(2n-1) = F(2n)
For instance, the sum of the first 6 odd terms is therefore F(2n) = F(12) = 144.
Fn^2 + F(n+1)^2 = F(2n+1)
F(n+1)F(n-1) - Fn^2 = (-1)^n
Some other relationships of Fibonacci numbers are Fn^2 + F(n+1)^2 = F(2n+1) and F(n+1)F(n-1) - Fn^2 = (-1)^n.
The ratio of the 2nth Fibonacci number divided by the nth Fibonacci number is always an integer or F2n/Fn = K. For instance, F10/F5 = 55/5 = 11.
The sum of any 4n consecutive Fibonacci numbers is evenly divisible by F2n.
Example: The sum of 4(2) = 8 Fibonacci numbers is divisible by F2(2) = F4 = 3.
1+2+3+5+8+13+21+34 = 87/3 = 29.
2+3+5+8+13+21+34+55 = 141/3 = 47.
3+5+8+13+21+34+55+89 = 228/3 = 76
As any number may be represented by combinations of powers of 2, so may any number be represented by combinations of the Fibonacci numbers.
1 = 1, 2 = 2, or 1+1, 3 = 3 or 1+2, 4 = 3+1, 5 = 5 or 3+2, 6 = 5+1 or 3+2+1, 7 = 5+2, 8 = 5+3, 9=5+3+1, 10 = 5+3+2, 50 = 34+13+3, 100 = 89+8+3 or 55+34+8+3, and so on.
The greatest common divisor of any two Fibonacci numbers is, itself, a Fibonacci number. Even more surprising is the fact that the g.c.d.(Fa, Fb) = c = F[g.c.d.(a,b)]. What this means is that the g.c.d. of the "a"th and "b"th Fibonacci numbers is the "c"th Fibonacci number where c = the g.c.d. of a and b.
For any two consecutive Fibonacci numbers, a and b, a^2 - ab - b^2 = +1 or -1.
Given the four consecutive Fibonacci numbers a, b, c and d, (ad), 2(bc) and (cd - ab) form a Pythagorean Triple such that (ad)^2 + [2(bc)]^2 = (cd - ab)^2.
Given two positive integers "m" and "n" that are relatively prime, the Fibonacci numbers F(m) and F(n) are also relatively prime.
An interesting aspect of Fibonacci like sequences.
Given the first 2 terms of a sequence where all successive terms derive from the recursive expression Fn = F(n-1) + F(n-2). The sum of any 10 consecutive terms of the sequence is the product of 11 times the 7th term of the chosen 10 terms.
Let the first term be "a" and the second term be (a + k). The 3rd term is then "a" + (a + k) = 2a + k. The 4th term is then (a + k) + 2a + k) = 3a + 2k. Continuing in this fashion
1. a
2. a + k
3. 2a + k
4. 3a + 2k
5. 5a + 3k
6. 8a + 5k
7. 13a + 8k
8. 21a + 13k
9. 34a + 21k
10. 55a + 34k (Note that the coefficients are the same as the terms of the Fibonacci sequence)
Adding, we derive the sum of S = 143a + 88k. Close examination of the sequence of terms shows that this sum is 11 times the 7th term of 13a + 8k.
The Fibonacci triangle
Sum of row and numbers above it Sum of row
1 1 = 1^2 1^3
3 5 9 = 3^2 2^3
7 9 11 36 = 6^2 3^3
13 15 17 19 100 = 10^2 4^3
21 23 25 27 29 225 = 15^2 5^3
etc.
The triangle is simply the consecutive listing of the odd numbers in triangular form.
The sum of the numbers in a row and all the numbers above it are squares.
The sum of the numbers in each row is the row number to the third power: 1, 8, 27, 64, 125, . . .
Each row has the row number of numbers in it. Row 1 has 1 number, row 2 has 2 numbers, row 3 has three numbers, etc.
The first number in each row can be found by adding 1 to the product of the row number (n) times the row number less 1 (n -1):
Row n: n(n-1) + 1 = n^2 - n + 1
Row 1: 1(1-1) + 1 = 1 - 1 + 1 = 1
Row 2: 2(2-1) + 1 = 4 - 2 + 1 = 3
Row 3: 3(3-1) + 1 = 9 - 3 + 1 = 7
Row 4: 4(4-1) + 1 = 16 - 4 + 1 = 13
Sources of additional information:
1--Number Theory by Oystein Ore, Dover Publications, Inc., 1988.
2--The Man Who Loved Numbers by Paul Hoffman, Hyperion, 1998
3--Numbers by Graham Flegg, Barnes & Noble Books, 1993.
4--Mathematical Gems I by Ross Honsberger, MAA, 1973
5--Great Moments in Mathematics Before 1650 by Howard Eves, MAA, 1983
6--The Book of Numbers by J.H. Conway and R.K. Guy, Copernicus, 1995
7--Lure of the Integers by Joe Roberts, MAA, 1992
8--A Mathematical Mosaic by R. Vakil, Brendan Kelly Publishing, Inc., 1996.
when the numerator and denominator of a fraction can only be divided evenly by one, the fraction has been what?
the formula dat u hav quoted is supposed 2b
Fn = [(1 + sqrt5)^n - (1 - sqrt5)^n]
..............(2^n)sqrt(5)
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this is so confusing
lol
mariam
xxx