Count the total number of subsets that don't have consecutive elements

Question

I'm trying to solve pretty complex problem with combinatorics and counting subsets. First of all let's say we have given set A = {1, 2, 3, ... N} where N <= 10^(18). Now we want to count subsets that don't have consecutive numbers in their representation.

Example

Let's say N = 3, and A = {1,2,3}. There are 2^3 total subsets but we don't want to count the subsets (1,2), (2,3) and (1,2,3). So in total for this question we want to answer 5 because we want to count only the remaining 5 subsets. Those subsets are (Empty subset), (1), (2), (3), (1,3). Also we want to print the result modulo 10^9 + 7.

What I've done so far

I was thinking that this should be solved using dynamical programming with two states (are we taking the i-th element or not), but then I saw that N could go up to 10^18, so I was thinking that this should be solved using mathematical formula. Can you please give me some hints where should I start to get the formula.

Thanks in advance.

Answer 1

Take a look at How many subsets contain no consecutive elements? on the Mathematics Stack Exchange.

They come to the conclusion that the number of non-consecutive subsets in the set {1,2,3...n} is the fib(n+2) where fib is the function computing the Fibonacci sequence for the number n+2. Your solution to n=3 conforms to this solution. If you can implement the Fibonacci algorithm, then you can solve this problem, but solving the question for a number as large as 10^18 will still be a challenge.

As mentioned in the comments here, you can check out the fast doubling algorithm on Hacker Earth.
It will find Fibonacci numbers in O(log n).