Proving a Problem involving Fibonacci numbers

Question

I'm working on proving the problem that states

$\text {The sequence}$ {$F_n$} $\text {is defined by the} \ F_1=F_2=1, F_{n+2}=F_{n+1}+F_n \ \text {for} \ n \ge 1.$ $\text {For any natural number m, define} \ v_2(m) \ as$ $v_2(m)=n \ if \ 2^n|m \ and \ 2^{n+1}|m.$ $\text {For all possible positive integer n that satisfy the equation:} $ $$v_2(n!)=v_2(F_1F_2F_3...F_n)$$

First, to start out I want to make sure that I'm understanding the problem $$v_2(5!)=v_2(1*1*2*3*5)$$ $$v_2(120)=v_2(30)$$ So am I supposed to see if $2^5|120$ or $2^5|30$

I guess I'm having trouble figuring out what my $m$ is supposed to be. Thanks for any and all the help you can give!

Answer 1

The $m$ on the left is $120$ and on the right is $30$. You are supposed to count the factors of $2$ in each and see if they are equal. As an example $v_2(48)=4$ because $2^4|48$ but $2^5 \not | 48$

Answer 2

I think you are expecting too much if you stay with the original definition given for any natural number m;that is v2(m) = n if 2^n and 2^(n+1) divide m. you then consider those integers satisfying v2(n!) = v2((F_1)(F_2).....(F_n)). In your example,we have v2(5!) = v2(120) = 2 since by definition 2^2|120 and 2^3|120,but no higher power of 2.For the right hand side,which involves the product of the first 5 Fibonacci numbers =30, we have by your definition that v2(30) = 0 since 2^0|30, 2^1|30,but no higher value of 2^k will work.Offhand, I see no eason that a factorial and product of Fibonacci numbers should have the same value of v2. Edwin Gray