Can someone please explain to me how the Fibonnaci numbers work?
484068
Example of Fibonnaci: 1,1,2,3,5,8,13,21\nEach number in the sequence is the sum of the 2 numbers before. I.E. 1+1=2; 1+2=3; 2+3=5; 3+5=8, etc.\n\nHope this helps.
351788
The Fibonacci series has a number of interesting properties.\n\nThe ratio of successive terms (term divided by next term) converges on 0.618 (the Golden Ratio). Get a calculator and fill in the gaps below\n\n5/8 = 0.625\n8/13 =\n13/21 =\n21/34 =\n34/55 =\n55/89 =\n\nAND\n\nThe ratio of successive terms (term divided by previous term) converges to 1.618 = the inverse of the Golden Ratio.\n\n8/5 = 1.6\n13/8 = 1.625\n21/13 =\n34/21 =\n55/34 =\n89/55 =\n\nThe numbers appear in nature eg in the arrangement of leaves on a stalk, sunflower heads, pine cones etc.\n\nThe generating formula, each term is the sum of the last two, can be used to generate other series e.g.\n\n1, 3, 4. 7, 11, 18, 29, 47, 76,\n\nCompare this to\n\n1, 2, 3, 5, 8, 13, 21, 34, 55\n\nand notice how the difference between 55 and 65 is 21, the differemce between 34 and 47 is 13, the difference between 21 and 29 is 8 etc.\n\nThe series starting 1, 4 and 1, 5 and 1, 6  show similar properties\n\n1, 6, 7, 13, 20, 33, 53, 86, 139\n1, 5, 6, 11, 17. 28, 45, 73, 118\n1, 4, 5, 9. 14, 23, 37, 60, 97, \n1, 3, 4. 7, 11, 18, 29, 47, 76,\n1. 2. 3, 5, 8, 13, 21, 34, 55\n\nand they all converge on the golden ratio too.\n\nOne can experiment with other generating formulae\n\neg each term = the sum of the 3 previous terms\n\n0, 0, 1, 1, 2, 4, 7, 13, 24, 44, 81\n\nor each term is twice the previous term plus the one prior to that\n\n0, 1, 2, 5, 12, 29, 70, 169\n\nand explore the properties of such series. Have fun!
673019
The golden ratio is approximately 1.618...\nFibonacci numbers are 1,1,2,3,5,8,13,21,44...\nOkay, now the relationship.\n1/1 is smaller than the golden ratio.\n2/1 is bigger than the golden ratio.\n3/2 is smaller than the golden ratio.\n5/3 is bigger than the golden ratio.\n8/5 is smaller than the golden ratio.\nand so on...
1004221
For the moment we don´t even know if there is an infinite number of Fibonacci primes....\n\nHere are the first fifteen:\n\n2,3,5,13,89,233,1597,28657,\n514229,433494437, 2971215073, 99194853094755497,\n1066340417491710595814572169,\n19134702400093278081449423917,\n475420437734698220747368027166749382927701417016557193662268716376935476241\n\nLooks considerably less dense than the fraction prime numbers / natural numbers
481505
The first three digits are actually multiplied by (80 x 250) = 20000. And you are adding 250 which you have subtracted latter.\n\nIn effect two times the first three digits followed by four zeros. You are adding two times the four digits so you get two times the original number. Dividing it by 2 u get the original value.\n\nYou can do this with 40 and 500, 20 and 1000, 10 and 2000 etc.
685370
you write the digits and then draw the bar above the repeating ones.  for example, if you wanted to write out 1 1/3 as a decimal, you would notice that it is 1.333333... so it would be written as 1.3 or (more often) 1.33, with a bar over only the "3"s.  if it is a number such as 2.345676767676... the bar would go above only the first set of "67," and not over the 3, 4, or 5, since they don't repeat.  some numbers may have only one repeating digit, and others may have many.  the bar goes over the first repetition only, and is not used in cases where there are other numbers that do not repeat interrupting the pattern
280927
let the first 3 digits of your number be x = abc and the last four digits be y =defg\n\nwhere a-g are the digits\n\nThe sequence of operations can be rewritten as:\n\n((x*80 + 1) * 250 + 2 * y - 250)/2\n\nIf you distribute the first 250 you get\n\n(20000x + 250 + 2y - 250)/2\n\nCancel the 250's\n\n(20000x+2y)/2\n\nDivide both elements of the numerator by two (distribute the 2)\n\n10000x + y\n\nSince x is a three digit number abc, abc * 10000 = abc0000\n\nSince y is a four digit number defg, 10000x + y = abcdefg.  \nThus we see using algebra that you end up with the same 7 digit number you started out with.
801584
If the number begins with 5, there are still three of the odd digits available to fill the last place.  If a 3 is used for the last place, the others are filled using 1, 2, 3, 4, and so there are 4! arrangements.\nIf 1 is used for the last place, that leaves 2, 3, 3, 4 for the others, and so the number of arrangements is  4!/2! because of the repeated 3.  So that gives \n4! + 4!/2! = 36\n\nIf it begins with a 3, the last place can be filled with 1, 3, or 5, and  the numbers left to fill the remaining four places are all different and so the number of possibilities is\n3*4! =72\n\nIf it begins with a 2 or 4 ( two choices), and the last with 1 or 5 (two choices), there are 4!/2! ways of filling the other places because 3 is repeated.\n2*2*4!/2! = 48\n\nIf it begins with 2 or 4 and ends with 3, then the other four digits are all different, and the number of possibilities is\n2*4! = 48\n\nAdding all these together gives 204.  Yes.
835878
There's a better way of looking at it.\n\nFor every real number x in [0,1) you define D(x, n) as the nth decimal of x.  This is defined for every natural number n.\n\nFor x=1/3, we have D(1/3, n) = 3 for every n.  The number 0.3333... (the dots are important, they indicate that the nth decimal is a 3 for every n) is not "very close" to 1/3, it does not "tend toward" 1/3.  It is EXACTLY the number 1/3.  Show me a number other than 1/3 and I will show you that its decimal representation is not 0.33333...\n\nFor 0, we have D(0,n) = 0 for every n.  There is never a '1' decimal.  If you give me an arbitrarily small number 0.00000...1, the digit 1 is at a finite position p.  I can then reply with a number such that D(x,n)=0 for all n<=p, and D(x,p+1) = 1.  That number is greater than zero and smaller than the number that you gave me.  \n\nThe 0.5 = 0.49999... thing is just a quirk of decimal notation.  We can formally prove the equality as follows:\n\nLet a = 0.5, i.e. D(a, 1) = 5 and D(a,n) = 0 for all n>2.\nLet b = 0.4999..., i.e. D(b, 1) = 4 and D(b,n) = 9 for all n>2.\n\nProof by contradiction that a=b: Let's suppose on the contrary that a > b.  This means there must be a number x between a and b.\n\nLet's figure out the digits of x.  Since x is less that a=0.5, its first decimal must be a 4.  D(x,1) = 4.\n\nThe second decimal of x must be 9 since x > b `> 0.49.\nThe third decimal of x must be 9 since x > b >0.499.\nAnd so on, the nth decimal of x must be a 9 because x > b > the number y such that D(y,1)=4, D(y,i)=9 for i=2 to n and D(y,i) = 0 for i>n.\n\nTherefore x has exactly the same digits as b=0.499999... (I just showed that for every n, the nth digit of x is the same as the nth digit of b), and so x=b.  Contradiction, ergo a = b.\n\nThat's the difference between 0.499999... vs. 0.5 and 0.0000...1 vs. 0.  In the latter case, it is possible to find a number that is between the two.
836949
With all good things like this you need to state the obvious first.\n\nA number divides by 3 if the remainder is zero if it divides by 3.\n\nFor numbers less than 10 this is obviously 3,6 and 9.\n\nLook at remainders of 10,20 and 30 when divided by 3. 1,2 and zero respectively.  Also notice that if you add two numbers together then to find the remainder when divided by three you only need to add remainders. So for example13 can be written as 10 + 3.  Looking only at remainders this is 1+0 =1 so the remainder when 13 is divided by 3 is 1.  You can reduce a number by 30 without affecting the remainder so higher numbers such as 50 can be reduced by 30 without affecting the remainder.  So you can see how numbers below 100 can be checked this way.\n\nNext notice that a number whose leading digit is one followed by zeros must have remainder 1 when divided by 3.  Take one from this number to get a digit whose numbers are all 9 and this is obviously divisible by 3.  This fact allows us to extend the reasoning we went through for 10, 20 etc to all higher numbers.\n\nThis explanation may be a bit hard going but hopefully it will be some sort of inspiration for your understanding.
26758
Why would it be 11, 10, 12?  It's 0, 1, 2, not 1, 0, 2.\n\nThe second digit (the 1 in 10) is the tens place, and the first digit (the 0 in 10) is the ones place, so after 9 you have no more digits to put a higher number in the ones place, so you put a 1 in the tens place and a 0 in the ones place and get 10.  To add one to that you put a 1 in the ones place and get 11.  Thus 11 is one more than 10 so it comes after it when you arrange the numbers from smallest to largest (which is essentially what you're doing when you're counting).
1065380
I won't give you the fomula- figure it out yourself.\nBelow are the possible combinations of 3 bit base 3; you decide on how a forth bit (used as a complement) would affect it.\n\n000\n001\n002\n010\n011\n012\n020\n021\n022\n100\n101\n102\n110\n111\n112\n120\n121\n122\n200\n201\n202\n210\n211\n212\n220\n221\n222
840916
i know how it's done. in the theorm, every number is one higher than stated (ex. 1=2, 2=3, 3=4, etc) since it's 2+2, it's actually 3+3. and if 3+3 =6, and 6=7, then the resulting equation is 2+2=7
671756
The base values are:\n\nI - 1\nV - 5\nX - 10\nL - 50\nC - 100\nD - 500\nM - 1000\n\nThese can be placed together so that consecutive numbers that are equal or smaller are added, but if a smaller number comes before a larger one, it is subtracted. So VI is 6, but IV is 4.\n\nLines can be placed over any of the letters to multiply them by 1000.
1142175
One property of fibonacci series is the following\n Take two consecutive fibonacci numbers and apply to them the euclidean division algorithm.\nThen they will get the maximum number of steps( or the maximum number of divisions).\n So a computer will require more time for division of big numbers if you use euclidean division algorithm.
916534
21, 34  Yes it's called the Fibonacci series.  Each number is the sum of the previous two.\n\nWhat I don't understand is as a result of asking a valid question to expand your knowledge, so many people gave you such rude replys.\n\nI hope this helps, have a good day.
945798
Most people use a base 10 system. Whatever you use, the base is that "Unreachable" number. For example, in base 10, there is no digit for 10, you start a new column representing how many 10's you've amassed. You have a 1's column a 10's column a 100's column and so forth. To get the numerical value you multiply (most people are so used to it, they just do it in their head). If the number is 823 you can express it as 3X1 plus 2X10 plus 8X100\n3x1 = 1\n2x10 = 20\n8X100=800\n-----------------\n1+20+800= 821\n\nSo 3 plus 20 plus 800. Make sense?\n\nIn binary "2" is the unreachable number, so you go 1, 10, 11, 100, etc. You have a 1's, 2's, 4's, 8's column.\nThe number 11011 would be 1Xone, plus 1xtwo, plus 0Xfour, plus 1xeight, plus 1x16 or 27 in base 10.\n\n1x 1 = 1\n1x 2 = 2\n0x 4 = 0\n1x 8 = 8\n1x16 = 16\n---------------\n1+2+8+16=27\n\nLet's look at your specific numbers 1011, that's \n\n1  x  1  =  1\n1  x  2  =  2\n0  x  4  =  0\n1  x  8  =  8\n--------------------\n1+2+0+8= 11\n\nYour top number is 11 (in base 10)\nYour bottom number:\n\n1 x 1 = 1\n1 x 2 = 2\n----------------\n1+2=3\n\nSo the problem states 11/3, that's pretty easy. 3.666(repeating)\n\nWith practice, it becomes easier. You learn to recognize the column values and mulitply in your head. \n\nHope that helps,\nOdd