Repeating fibonacci sequence formula8/15/2023 ![]() ![]() so we can calculate the sum for very large values. Nice thing about this is that its an O(1) (or O(log(N)) if you include the cost of pow/log) algorithm, and works on doubles. The sum of all even numbers is half of this: sum_even = 4613732 The sum of all Fibonacci numbers up to this point if given by sum = floor( pow( phi, n+2 )/sqrt(5) + 0.5 ) - 1 = 9227464 int n = floor(log(4000000*sqrt(5))/log(phi)) = 33.ģ3 is divisible by 3 so it is an even Fibonacci number, if it wasn't we'd need to adjust n like this. ![]() So convert our maximum value of 4000000 calculate the index of the highest Fibonacci number The sum of the even Fibonacci numbers is half the sum of the odd Fibonacci numbers upto the same point in the sequence.( Look at the sequence mod 2 and the result is trivial ) Any (generalized) Fibonacci sequence modulo m must repeat. The even Fibonacci numbers are are every third number. sequence, it would be nice to know a formula for Fn so we wouldnt have to compute all the.The sum of the first N Fibonacci numbers is the N+2th fibonacci number minus 1 (See Second Identity on the same wikipedia page).Conversely the index of the biggest Fibonacci number less than i is given by floor( log(i*sqrt(5)) / log(phi) ) Each Fibonacci number can be calculated directly as floor( pow( phi, n ) + 0.5 ) (See Computation by Rounding in ).Here's an alternate (and very fast, but untested, method). NOTE: This has been heavily modified to address the actual question I hope this shows that a functional programming approach is not difficult and is a more elegant way to solve certain types of problem. ![]() To solve the problem you can use sum: sum(list(takewhile(lambda x: x < 4000000, evenfibs))) To fetch numbers from the generator until a number exceeds four million use takewhile: for x in takewhile(lambda x: x < 4000000, evenfibs): To find only the even numbers we can use ifilter to produce a new generator: def is_even(x): To print the first few numbers use islice: from itertools import ifilter, islice, takewhile To use this generator we can import a few useful functions from itertools. But Python also supports functional programming to a certain extent and for this sort of exercise a functional programming approach is more enlightening in my opinion.įirst define a generator that generates all the Fibonacci numbers: def fib(): This is not surprising because Python is an imperative language. Most people when they start learning Python learn only imperative programming. You have repeated code in the body of your loop.You are looping four million times instead of until a condition is true.There are a couple of problems with your code: Working code (finished in less than one second): Fibonacci1 = 1 Fn Fn-1+Fn-2 Here, the sequence is defined using two different parts, such as kick-off and recursive relation. I though it means that i have to sum all the even terms UP TO 4 million ! But the solution was to sum all the even terms UNTIL 4 million. It should takes less than a minute, but it took all night and it wasn't solved !Įdit: Sorry guys, i misunderstood the problem. Hey, i was trying to write a script which sums all the even terms in "Fibonacci Sequence" under 4 millions. These numbers, 34 and 21, are numbers in the Fibonacci series, and their ratio 1.6190476 closely approximates Phi, 1.6180339.Python program to find fibonacci series. The DNA molecule measures 34 angstroms long by 21 angstroms wide for each full cycle of its double helix spiral. DNA moleculesĮven the microscopic realm is not immune to Fibonacci. When a hawk approaches its prey, its sharpest view is at an angle to their direction of flight - an angle that's the same as the spiral's pitch. And as noted, bee physiology also follows along the Golden Curve rather nicely. Following the same pattern, females have 2, 3, 5, 8, 13, and so on. Thus, when it comes to the family tree, males have 2, 3, 5, and 8 grandparents, great-grandparents, gr-gr-grandparents, and gr-gr-gr-grandparents respectively. Males have one parent (a female), whereas females have two (a female and male). In addition, the family tree of honey bees also follows the familiar pattern. The answer is typically something very close to 1.618. The most profound example is by dividing the number of females in a colony by the number of males (females always outnumber males). Speaking of honey bees, they follow Fibonacci in other interesting ways. ![]()
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