# fibonacci series - a slight variant

• Jul 8th 2010, 01:01 AM
aman_cc
fibonacci series - a slight variant
Let me define a recurence like this

S(1) = 1
S(2) = 1

S(n) = S(n-1) + S(n-2) + a; where a is some number

Q1: What is the relation beween S(n) and F(n) [nth Fibonacci number]

I found out this is S(n) = F(n) + K(n)*a

Where K(n) is another series of the type S(n) where a = 1

Q2: Is there a way to find the generic term for S(n) ?

Any pointers will be of great help. Thanks
• Jul 8th 2010, 01:25 AM
Ackbeet
Well, you can always use the theory of difference equations (recurrence relations). You can find the nth term of the Fibonacci sequence this way, and I'm sure your S(n) sequence can be found the same way. The theory is very similar to that of differential equations.
• Jul 8th 2010, 02:07 AM
simplependulum
Let $f(n)$ be a sequence that $f(n+2) = f(n+1) + f(n)$ , not necessarily be Fibonacci seq. , depends on the initial conditions .

Then we have $S(n) = f(n) - a$ because

$f(n+2) - a = f(n+1) + f(n) - a = (f(n+1) - a) + (f(n) - a ) + a$

$= S(n+1) + S(n) + a$
• Jul 8th 2010, 07:20 AM
chiph588@
Quote:

Originally Posted by aman_cc
Let me define a recurence like this

S(1) = 1
S(2) = 1

S(n) = S(n-1) + S(n-2) + a; where a is some number

Q1: What is the relation beween S(n) and F(n) [nth Fibonacci number]

I found out this is S(n) = F(n) + K(n)*a

Where K(n) is another series of the type S(n) where a = 1

Q2: Is there a way to find the generic term for S(n) ?

Any pointers will be of great help. Thanks

We have $S_3=2+a, \; S_4=3+2a$.

You can use induction to show $S_n=F_n+a\cdot (F_n-1)$.

Edit: Credit goes to Soroban's post below. I had made a mistake.
• Jul 8th 2010, 08:01 AM
Soroban
Hello, aman_cc!

Quote:

Let me define a recurence like this:

. . $S(1) = 1$
. . $S(2) = 1$
. . $S(n) \:= \:S(n\!-\!1) + S(n\!-\!2) + a\:\text{ for some constant }a.$

What is the relation beween $S(n)$ and $F(n)$, the $n^{th}$ Fibonacci number?

Crank out the first few terms:

. $\begin{array}{|c||c|c|}n & S(n) & F(n)\\ \hline
1 & 1 & 1\\
2 & 1 & 1\\
3 & 2+a & 2 \\
4 & 3+2a & 3\\
5 & 5+4a & 5 \\
6 & 8 + 7a & 8 \\
7 & 13 + 12a & 13 \\
8 & 21 + 20a & 21 \\
\vdots & \vdots & \vdots
\end{array}$

And we see that: . $S(n) \;=\;F(n) + [F(n)-1]\cdot a$

• Jul 12th 2010, 02:59 AM
aman_cc
Thanks for all the posts