There are multiple straight-forward ways to solve the given recurrence like using Generating Functions, Linear Algebra or you can write the characteristic equation for the given recurrence with constant coefficients and find the roots and get the closed-form expression with given conditions and evaluate $F_1.$

But it can be solved by the following two other methods and you can generalize it also if you have given like $F_{1000},F_{1001}$ and ask about $F_1.$

**Method 1:**

Zeckendorf’s theorem tells that “every positive integer can be represented uniquely as the sum of *one or more* distinct Fibonacci numbers in such a way that the sum does not include any two consecutive Fibonacci numbers.”

Here, given recurrence is same as we use for Fibonacci numbers but different values as stated in the question.

As Fibonacci numbers $f_0,f_1,f_2,f_3,f_4,f_5,f_6,f_7,f_8,f_9,f_{10},..$ are $0,1,1,2,3,5,8,13,21,34,55,...$ (if you start the Fibonacci numbers from zero)

Here, you are given $F_7 = 60=55+5 = f_{10}+f_5$ and similarly,

$F_6 = 37=34+3 = f_{9}+f_4$ and so,

$F_5 =F_7 -F_6 = (f_{10} \ – \ f_9) + (f_5 -f_4) = f_8 + f_3$

Hence, in General, you can write:

$$F_i = f_{i+3} + f_{i-2}, i \geq 2$$

So, $F_2= f_5+f_0 = 5+0 = 5$ and $F_3= f_6+f_1 = 8+1 = 9$

Therefore, $F_1 = F_3 \ – \ F_2 = 9-5 =4$

In General, if you have given a very large positive integer, then you can find the nearest Fibonacci number by using this and write that integer with sum of one or more distinct Fibonacci numbers and then compute $F_i$ for some integer $i.$