Muhammad's Tri-onacci subsequence

For any $n$ we can use the recurrence relation for F to find:

$F_{n+2} = F_{n} + F_{n+1}$

$F_{n+3} = F_{n} + 2F_{n+1}$

$F_{n+4} = 2F_{n} + 3F_{n+1}$

$F_{n+5} = 3F_{n} + 5F_{n+1}$

$F_{n+6} = 5F_{n} + 8F_{n+1}$

Inspecting this list, we see that $F_{n+6} = 4F_{n+3} + F_{n}$

so the required coefficients are $4$ and $1$ .

The recurrence given is equivalent to $$F {3n} = F {3n - 3} + F {3n - 6}$$ so we have $$F 6 = aF 3 + bF 0$$ $$F 9 = aF 6 + bF_3$$ i.e. $$8 = a\cdot 2 + b \cdot 0$$ $$34 = a\cdot 8 + b \cdot 2$$ Solving this system of linear equations yields $a = 4$ , $b = 1$ , so $$a + b = \boxed{5}$$

We know that, $a$ and $b$ are constants, so for any value of $G_{N}$ , the values of $a$ and $b$ will be same.
Here $G_{n}$ = $F_{3n}$ , $G_{n-1}$ = $F_{3n-3}$ , $G_{n-2}$ = $F_{3n-6}$
When, $n$ = $3$ , then,
$F_{9}$ = $a$ $F_{6}$ + $b$ $F_{3}$ = $34$ [ We can easily count $F_{9}$ = $34$ ]
⇒ $8$ $a$ + $2$ $b$ = $34$ ....................(1)
When, $n$ = $4$ , then,
$F_{12}$ = $a$ $F_{9}$ + $b$ $F_{6}$ = $144$ [ We can easily count $F_{12}$ = $144$ ]
⇒ $34$ $a$ + $8$ $b$ = $144$ ....................(2)
By solving the equation (1) and (2), we shall get $a$ = $4$ and $b$ = $1$
So, $a$ + $b$ = $4$ + $1$ = $5$ [ ANSWER ]

A quick solution is to write out part of $G$ ; we get $0,2,8,34,144,\cdots$ We know $8$ must equal $2a+0b$ , so $a=4$

We now know that $34=4\times8+2b=32+2b \rightarrow b=1$ , thus $a+b=4+1=\boxed{5}$

A more elegant solution uses what we know about $F_n$ and $G_n$ . We can create an equation $G_n=F_{3n}=F_{3n-1}+F_{3n-2}$ . Thus, $G_{n-1}=F_{3n-4}+F_{3n-5}$ and $G_{n-2}=F_{3n-7}+F_{3n-8}$ . Next we can simplify $G_n$ recursively into $\underbrace{F_{3n-7}+F_{3n-7}+\cdots+F_{3n-7}}_{21}+\underbrace{F_{3n-8}+F_{3n-8}+\cdots+F_{3n-8}}_{13}$ and $G_{n-1}$ into $\underbrace{F_{3n-7}+F_{3n-7}+\cdots+F_{3n-7}}_5 +\underbrace{F_{3n-8}+F_{3n-8}+F_{3n-8}}_3$ So we need $a$ and $b$ so that $21=5a+b$ and $13= 3a+b$

Linear combination gives $a=4,b=1 \rightarrow a+b=4+1=\boxed{5}$

Let us take a look at the first $12$ numbers of the Fibonacci sequence

$a_n=1,1,2,3,5,8,13,21,34,55,89,144\ldots$

The third numbers in this segment of the sequence are $2,8,34,144$ .

According to the problem, $G_{n}=aG_{n-1}+bG_{n-2}$ .

Using the "third" numbers we listed above we get the system

$\left\{ \begin{array}{l l} 8a+2b=34 & \quad \\ 34a+8b=144 & \quad \end{array} \right.$

By solving it, we find that $a=4$ and $b=1$ , therefore

$a+b=5$ .

Since we know the constants exists, we can just write the equation for the particular cases G3 and G2 (we have a system of two liner equations with two unknowns). It easily follows a = 4 and b = 1.

f(3n) =f(3n-1)+f(3n-2) =f(3n-2)+f(3n-3)+f(3n-3)+f(3n-4) =f(3n-3)+2 (f(3n-4)+f(3n-4))+f(3n-5)+f(3n-5)+f(3n-6) reordering them =f(3n-3)+(f(3n-4)+f(3n-5)) 3+f(3n-6) =4*f(3n-3)+f(3n-6)

After listing out all Fibonacci numbers up to $F_{15}$ , we can consequentially determine $G_2, G_3, G_4,$ and $G_5$ .

$G_2 = F_6 = 8$ $G_3 = F_9 = 34$ $G_4 = F_{12} = 144$ $G_5 = F_{15} = 610$

Given the definition of $G_n$ , we can establish two equations:

$G_4 = 144 = 34a + 8b$ $G_5 = 610 = 144a + 34b$

Solving for $a$ and $b$ , we get 4 and 1, respectively. Thus $a + b = 4 + 1 = 5$ .

0,1,1,2,3,5,8,13,21,34 34=a(8)+b(2)

G 0 = 0, G 1 = 2, G 2 = 8, G 3 = 34. Then 8 = a * 2 + b* 0, and that impose a = 4. Therefore 34 = 4 * 8 + b * 2. One obtain b = 1. Then a + b = 4 + 1 = 5.

$F_{3n+6}=aF_{3n+3}+bF_{3n}$ $F_{3n+6}=F_{3n+5}+F_{3n+4}=F_{3n+4}+F_{3n+3}+F_{3n+3}+F_{3n+2}$ $=F_{3n+3}+F_{3n+2}+F_{3n+3}+F_{3n+3}+F_{3n+1}+F_{3n}$ $=3F_{3n+3}+F_{3n+2}+F_{3n+1}+F_{3n}$ $=4F_{3n+3}+F_{3n}$

From the definition of the Fibbonacci sequence:

$F_{3n}=aF_{3(n-1)}+bF_{3(n-2)}$ for all $n\geq2$

$$

Let $n=2$ :

$F_{6}=aF_{3}+bF_{0}$

$F_{6}=8,F_{3}=2,F_{0}=0$

$8=2a+0b$

$a=4$

$$

Let $n=3$ :

$F_{9}=4F_{6}+bF_{3}$

$F_{9}=34,F_{6}=8,F_{3}=2$

$34=4*8+2b$

$2b=2$

$b=1$

$$

$a+b=4+1=5$

G0 = F0 = 0,
G1 = F3 = 1, G2 = F6 = 5, G3 = F9 = 21 and so on... so we get G sequence as 0, 1, 5, 21, 89... now for n >= 2, from the given equation, G2 = aG1 + bG0 = 5 , G3 = aG2 + bG1 = 21 , G4 = aG3 + bG2 = 89 , we can see a pattern here like 5 = 4(1) + 1 , 21 = 4(5) + 1(1) , 89 = 4(21) + 1(5) , comparing this with the previous equations, we get a=4 and b=1

F(3n)=F(3n-1)+F(3n-2)

G(n)=F(3n-2)+F(3n-3)+F(3n-3)+F(3n-4)

G(n)=F(3n-3)+F(3n-4)+2xF(3n-3)+F(3n-4)

G(n)=3xF(3n-3)+2xF(3n-4)

G(n)=3xG(n-1)+2x[F(3n-5)+F(3n-6)]

G(n)=3xG(n-1)+F(3n-4)+F(3n-5)+F(3n-5)+F(3n-6)+F(3n-6)-F(3n-4)

G(n)=3xG(n-1)+F(3n-3)+F(3n-6)

G(n)=4xG(n-1)+G(n-2)

Hence a+b is 5

First, we have G(0) = 0; G(1) = 2; G(2) = 8; G(3) = 34; G(4) = 144 .... Then, we can predict that G(n+2) = G(n+1) + 4.G(n) I think it's easy to prove this equation by inductive method.

Since we know that Gn = a(Gn-1) + b(Gn-2) we can express this as a sistem of two linear equations. They will be: 34(which is the 9th number of the Fibonacci sequence) = 8a(8 is equal than the 6th number of the Fibonacci sequence) + 2b(2 is equal than the 3th number of the Fibonacci sequence). We can do the same for higher numbers in Fibonacci sequence and we will get that: 144(12th number) = 34a(9th number) + 8b(6th number). Solving this linear equations we get: a = 4 and b = 1. And so, a + b = 5 SOLVED

G1=f3=2, G2=f6=8, G3=f9=34=8a+2b, G4=f12=144=34a+8b. From the first and second equation we get a=4 and b=1. So, a+b=5

G0=0, G1=2, G2=8, G3=34. G2=aG0+bG1, 8=2b. G3=aG1+bG2, 34=2a+8b. Solving the simultaneous equations, a=1, b=4, a+b=5.