# Computing nth term of fibonacci-like sequence for large n

Sum up to nth term of fibonacci sequence for very large n can be calculated in O($\log n$) time using the following approach:

$$A = \begin{bmatrix} 1&1 \\\\1&0\end{bmatrix}^n$$ $$\begin{bmatrix}f(n+1) \\\\ f(n) \end{bmatrix} = A^n \begin{bmatrix}f(1) \\\\ f(0)\end{bmatrix}$$

We can calculate $A^n$ in O($\log n$) time by calculating $A$, $A^2$, $A^4$, $A^8$...

Now I have another sequence

$$T(n) = T(n - 1) + T(n - 2) - (4n - 13)$$ $$T(1) = 3$$ $$T(2) = 8$$

I want to calculate its nth term for large n in O($\log n$) time.

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Let $T(n)=S(n)+an+b$, where $a,b$ will be decided later...

Then

$$S(n)+an+b=S(n-1)+an-a+b +S(n-2)+an-2a+b -(4n-13)$$

Thus

$$S(n)=S(n-1)+S(n-2) +an-3a+b -(4n-13) \,.$$

Now, if we can make

$$an-3a+b=4n-13 \,, (*)$$

we get

$$S(n)=S(n-1)+S(n-2) \,,$$

and hence,as in Fibonnaci,

$$\begin{bmatrix}S(n+1) \\\\ S(n) \end{bmatrix} = A^n \begin{bmatrix}S(1) \\\\ S(0)\end{bmatrix}$$

you can now calculate $S(n)$ within $O(\log n)$ time, and to get $T(n)$ you need to add $an+b$, where $a,b$ are calculated from $(*)$.

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Thanks a lot! This was very helpful. –  Aayush Bahuguna Oct 10 '12 at 22:35
It is a well known technique, basically whenever when you have a linear recurrence + extra stuff, you want to add something to $t(n)$ which cancels the extra stuff... This technique is also used in differential equations, I think it is called the method of undetermined coefficients.... –  N. S. Oct 10 '12 at 22:39
@RossMillikan I canceled one $an$, note that there were 2 on LHS and one on RHS ;) –  N. S. Oct 10 '12 at 22:55
Is the same procedure(comparing coefficients) applies for the sequence having extra stuff which is not linear i.e T(n) = T(n - 1) + T(n - 2) + P(n) where, P(n) is degree d polynomial. Is there any easy method for this kind of equation? –  Aayush Bahuguna Oct 11 '12 at 17:21
@AayushBahuguna If it is $T(n)=T(n-1)+T(n-2)+P(n)$, you can always find a polynomial $Q(n)$ of the same degree as $P$ so that $S_n=T(n)-P(n)$ satisfies the Fibonacci equation. –  N. S. Oct 11 '12 at 20:26

Let $S(n) = T(n) - 4n - 25$, then \begin{align*} S(n) &= T(n) - 4n - 25 \\&= T(n-1) + T(n-2) - 8n - 13 - 25 \\&= S(n-1) + 4n - 4 + 25 + S(n-2) + 4n - 8 + 25 - 8n - 13 - 25 \\&= S(n-1) + S(n-2) \end{align*} So $S(n)$ fulfills the Fibonacci recurrence. Now do as above and calculate $T(n) = S(n) + 4n+25$ afterwards.

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@Ross thx, will edit. –  martini Oct 11 '12 at 8:22