The Distribution of Fibonacci Numbers Modulo PrimesDistributionFibonacci Numbers Modulo PrimesShah-Bruckner theoremWe give a new proof of the Shah–Bruckner theorem, which states that | Ω p | 7, where Ω p is the set of Fibonacci numbers modulo p . We also show that lim inf p →∞ | Ω...
1≡αx−p+βx−q(modx2−x−1). To get the actual solution from it, we should first understand what exactly is the remainder of xnxn modulo x2−x−1x2−x−1. The remainder of P(x)P(x) modulo (x−a)(x−b)(x−a)(x−b) is generally determined by P(a)...
#P193E. Fibonacci Number Description John Doe has a list of all Fibonacci numbers modulo1013. This list is infinite, it starts with numbers0and1. Each number in the list, apart from the first two, is a sum of previous two modulo1013. That is, John's list is made from the Fibonacci ...
In mathematical terms, the sequenceFnof Fibonacci numbers is defined by the recurrence relation F1 = 1;F2 = 1;Fn = Fn - 1 + Fn - 2(n > 2). DZY loves Fibonacci numbers very much. Today DZY gives you an array consisting ofnintegers:a1, a2, ......
The Fibonacci numbers are defined as below: Given three integers N, C and K, calculate the following summation: Since the answer can be huge, output it modulo 1000000009 (10^9+9). Input The first line contains an integer T (1≤T≤200), denoting the number of test cases. Each test ca...
The Fibonacci numbers are defined as below: Given three integers N, C and K, calculate the following summation: Since the answer can be huge, output it modulo 1000000009 (10^9+9). Input The first line contains an integer T (1≤T≤200), denoting the number of test cases. Each test ca...
Based on the approach in my previous blog, today, I found an amazing way to calculate large fibonacci numbers (in some modulo). According to part IV of my previous blog, let f(n) be the (n + 1)th fibonacci number, we have two case: n is even and n is odd. f(2 * ...
The Fibonacci numbers are defined as below: F_0=1,F_1=1\\ F_n=F_{n-1}+F_{n-2}\ (n>1)\\ Given three integers N , C and K , calculate the following summation: (F_0)^K+(F_C)^K+(F_{2C})^K+...+(F_{NC})^K\\ Since the answer can be huge, output it mo...
Thus, considering two arbitrary consecutive members of the sequence modulo j , i.e the ordered pairs (Fn (mod j ), Fn+1 (mod j )), we see that there are j choices for each Fn and Fn+1 so that there are j 2 possibilities for these consecutive numbers. Since there are a ?nite ...
modulo1000000009 (109 + 9). Help DZY reply to all the queries. Input The first line of the input contains two integersnandm(1 ≤ n, m ≤ 300000). The second line containsnintegersa1, a2, ..., an(1 ≤ ai ≤ 109)— initial arraya. ...