1. ## Sequences

I can't seem to figure out what the next two terms of this sequence are as I can't figure out the term-to-term rule for it. Can anyone help?

11, 9, 10, 17, 36, 79,....

2. Originally Posted by nsingh201
I can't seem to figure out what the next two terms of this sequence are as I can't figure out the term-to-term rule for it. Can anyone help?

11, 9, 10, 17, 36, 79,....
I don't know if this helps you at all, but I believe it is possible to find a polynomial of at most degree 5 that will fit that sequence, by solving simultaneous linear equations.

I will illustrate the method on a smaller scale. Say the sequence is

11, 9, 10, ...

Then we can find a polynomial of degree 2 that will generate this sequence, of the form f(x) = ax^2+bx+c, as follows:

We know f(1) = 11. Thus, a*(1^2)+b*1+c = 11. Simplifying, a+b+c=11.

We know f(2) = 9. Thus, a*(2^2)+b*2+c = 9. Simplifying, 4a+2b+c=9.

We know f(3) = 10. Thus, a*(3^2)+b*3+c = 10. Simplifying, 9a+3b+c=10.

This is a system of linear equations with three equations and three unknowns. The solution is described by (a,b,c) = (3/2, -13/2, 16). That is, f(x) = (3/2)x^2 - (13/2)x + 16. This generates the sequence

11, 9, 10, 14, 21, 31, ...

Obviously this is not the desired sequence, so a quadratic curve-fitting scheme is not feasible. I would proceed by next trying a cubic polynomial to model the first four terms, and if that doesn't work, a degree 4 polynomial to include the first five terms, and in the worst case, a degree 5 polynomial.

Of course there may be a much simpler explanation of your sequence that I'm blind to...

3. 11,9,10,17,36,79,170,357,736,1499,3030,6097,12236, 24519,...

4. ## ?

5. Hello, nsingh201!

Find the next two terms of this sequence: . $11,\: 9,\: 10,\: 17,\: 36,\: 79 \;\hdots$

Given a sequence of increasing terms, I take the differences of consecutive terms,
. . then the differences of the differences, and so on . . . hoping to find a pattern.

Here's what I found . . .

. . $\begin{array}{ccccccccccccc}
\text{Sequence} & 11 && 9 && 10 && 17 && 36 && 79 \\
\text{1st diff.} &&\text{-}2 && 1 && 7 && 19 && 43 \\
\text{2nd diff.} &&& 3 && 6 && 12 && 24 \\
\text{3rd diff.} &&&& 3 && 6 && 12 \end{array}$

Since the 3rd differences seem to be identical to the 2nd differences,
. . the function appears to be exponential.

I also suspect that the 2nd differences are doubling.

So I believe the chart can be continued like this;

. . $\begin{array}{cccccccccccccccccc}
\text{Sequence} & 11 && 9 && 10 && 17 && 36 && 79 && {\color{red}170} && {\color{red}357} \\
\text{1st diff.} &&\text{-}2 && 1 && 7 && 19 && 43 && {\color{blue}91} && {\color{blue}187} \\
\text{2nd diff.} &&& 3 && 6 && 12 && 24 && {\color{blue}48} && {\color{blue}96}\end{array}$

And I think the generating function is: . $f(n) \;=\;3\!\cdot\!2^{n-1} - 5n + 13$

6. Originally Posted by Soroban
Hello, nsingh201!

Given a sequence of increasing terms, I take the differences of consecutive terms,
. . then the differences of the differences, and so on . . . hoping to find a pattern.

Here's what I found . . .

. . $\begin{array}{ccccccccccccc}
\text{Sequence} & 11 && 9 && 10 && 17 && 36 && 79 \\
\text{1st diff.} &&\text{-}2 && 1 && 7 && 19 && 43 \\
\text{2nd diff.} &&& 3 && 6 && 12 && 24 \\
\text{3rd diff.} &&&& 3 && 6 && 12 \end{array}$

Since the 3rd differences seem to be identical to the 2nd differences,
. . the function appears to be exponential.

I also suspect that the 2nd differences are doubling.

So I believe the chart can be continued like this;

. . $\begin{array}{cccccccccccccccccc}
\text{Sequence} & 11 && 9 && 10 && 17 && 36 && 79 && {\color{red}170} && {\color{red}357} \\
\text{1st diff.} &&\text{-}2 && 1 && 7 && 19 && 43 && {\color{blue}91} && {\color{blue}187} \\
\text{2nd diff.} &&& 3 && 6 && 12 && 24 && {\color{blue}48} && {\color{blue}96}\end{array}$

And I think the generating function is: . $f(n) \;=\;3\!\cdot\!2^{n-1} - 5n + 13$

beautiful approach

7. 11, 9, 10, 17, 36, 79, - Wolfram|Alpha
2 Soroban
Nice solution, but how did u get generating function?

8. Hello, ICanFly!

I was afraid someone would ask . . .LOL!

. . .but how did u get generating function?

We have: . $11,\:9,\:10,\:17,\:36,\;\hdots$

And we know there is a "doubling" feature in there: . $2^n$

Since the doubling didn't begin until the second differences.
. . I assumed there was a linear or quadratic involved, too.

So, my first general function was: . $f(n) \:=\:A + Bn + C\!\cdot\!2^n$

I used the first three terms of the sequence:

. . $\begin{array}{ccccc}
f(1) = 11 & A + B + 2C &=& 11 & [1] \\
f(2) \:=\: 9\; & A + 2B + 4C &=& 9 & [2] \\
f(3) = 10 & A + 3B + 8C &=& 10 & [3] \end{array}$

$\begin{array}{ccccc}
\text{Subtract [2] - [1]:} & B + 2C &=& -2 & [4] \\
\text{Subtract [3] - [2]:} & B + 4C &=& 1 & [5] \end{array}$

$\text{Subtract [5] - [4]: }\;\;2C \:=\: 3 \quad\Rightarrow\quad C \:=\:\tfrac{3}{2}$

Substitute into [4]: . $B + 3 \:=\:-2 \quad\Rightarrow\quad B \:=\:-5$

Substitute into [1]: . $A - 5 + 3 \:=\:11 \quad\Rightarrow\quad A \:=\:13$

Hence: . $f(n) \;=\;13 - 5n + \tfrac{3}{2}\!\cdot\!2^n$

I tested my function up to the 8th term . . . It works!

. . Therefore: . $f(n) \;=\;3\!\cdot\!2^{n-1} - 5n + 13$

9. Beautiful

10. thanks so much. this all really helped.