fibonacci

This logic programming is really making a lap dance on my imperative programming skills. This is homework, so please just don't drop me the answer. This is what I have: fibo(N,1) :- N < 2, !. fibo(N,R) :- N1 is N-1, N2 is N-2, fibo(N1,R1), fibo(N2,R2), R is R1+R2. I'm suppose to make another function that looks like this; fib(N,Value,LastValue) . N is the n'th number, and value is the return value. I don't understand how I can rewrite this using accumulation. And since it counts backwards I don't see how it can "know" a last value before it calculates anything. :s Any input is appreciated. I

Determining the Fibonacci sequence is easy enough to figure out: int num = 0; int num2 = 1; int loop; int fibonacci; System.out.print(num2); for (loop = 1; loop <= 10; loop ++) { fibonacci = num + num2; num = num2; num2 = fibonacci; System.out.print(" " + fibonacci); } My problem lies with trying to pinpoint the value for a specified N. As in, If I want to find the 6th element in the sequence, which is 8, how would I find that number, and only that number? In your code, num starts as the 0 th Fibonacci number, and num1 as the 1 st . So to find the n th , you have to iterate the step n times:

I am calculating the n-th fibonacci number using (a) a linear approach, and (b) this expression Python code: 'Different implementations for computing the n-th fibonacci number' def lfib(n): 'Find the n-th fibonacci number iteratively' a, b = 0, 1 for i in range(n): a, b = b, a + b return a def efib(n): 'Compute the n-th fibonacci number using the formulae' from math import sqrt, floor x = (1 + sqrt(5))/2 return long(floor((x**n)/sqrt(5) + 0.5)) if __name__ == '__main__': for i in range(60,80): if lfib(i) != efib(i): print i, "lfib:", lfib(i) print " efib:", efib(i) For n > 71 I see that the

Ok, I initially wrote a simple code to return the Fibonacci number from the series based on the user input.. n=5 will produce 3.. static int fibonacci(int n) { if (n == 1) return 0; else if (n == 2) return 1; else return (fibonacci(n - 1) + fibonacci(n - 2)); } I was thinking of modifying the code to return the sum of the series rather than just returning the value from the series and while trying to do the sum I accidentally added 1 to the return statement and to my surprise, it returned the sum correctly. The below code will return 7 for n=5. I am not sure if this is a right way to calculate