How do I find the rule for this number sequence? 7,21,42,126,252,756

How do I find the rule for number sequence: 7,21,42,126,252,756

The numbers are multiplied by 3, then by 2, by 3, then by 2, and so on. The next number would be 1512, then 4536, etc.

galactus, how do you find the sequence?
Trial and error? Look and see what works?

galactus, how do you find the sequence?
Trial and error? Look and see what works?

Finding the sequence is easy.

7 x 3 = 21

21 x 2 + 42

42 x 3 = 126

It's common repetition. If you look closely at all of the numbers in the list, you can see a pattern.

galactus, how do you find the sequence?
Trial and error? Look and see what works?

No, I did not use trial and error. I just 'seen' it.

No, I did not use trial and error. I just 'seen' it.

Me too, and I've been awake for ever. I guess it's just something you get used to.

Yep, just seeing '7', '21', and '42', we know they are multiples of 7. So, there could be some multiplication in there.

How about a sequence like:
1, 4, 9, 17, 34, 76, 99?

The above sequence are random numbers, Basically, I'm asking, is their a Formula, or programs that can determine, or do you just see what works?

How about a sequence like:
1, 4, 9, 17, 34, 76, 99?

Ok, I have two options, either that is not a sequence, or the formula is very complex.

If plotted, there are no real pattern. The 'increase' decreases after 76, which is quite a rare sight to see.

EDIT: Oops, seems that you posted before I did. I checked here, then had to give food to my dog and came back to answer.

The above sequence are random numbers, Basically, I'm asking, is their a Formula, or programs that can determine, or do you just see what works?

There is no said formula that determines the pattern of all sequences in general.

You just see the pattern for each sequence.

Some are easier than others.

Did you just make that one up? It is not a sequence with any definitive pattern that I can see. That does not mean there is not one though. I am not that arrogant:):)

One thing that can often be done is what is called 'finite differences'.

Using this method one can find a quadratic, cubic, etc that will give the nth term of the sequence, assuming there is one.

Let's say we have:

1 \;\ 4 \;\ 10 \;\ 20 \;\ 35 \;\ 56

Subtract between the numbers and we see a pattern:

3 \;\ 6 \;\ 10 \;\ 15 \;\ 21

Keep going:

3 \;\ 4 \;\ 5 \;\ 6

1 \;\ 1\;\ 1

See? We ended up with a bunch of 1's.

Since we have 3 rows before we got to the 1's, the polynomial that represents it will be a cubic.

I am not going to show all the work, but suffice to say it is:

\frac{1}{6}x^{3}+\frac{1}{2}x^{2}+\frac{1}{3}x

Now, we can plug in any number for x and find the number in the sequence.

Say we want the 10th number in the sequence.

\frac{1}{6}(10)^{3}+\frac{1}{2}(10)^{2}+\frac{1}{3 }(10)=220

I used a simple example. We can see these are the triangular numbers and find the next one without a polynomial, but if we wanted a number further up the ladder it is handy.

Another sequence is called a 'recursion relation'.

Say we had

a_{n}=7a_{n-1}-10a_{n-2}, \;\ a_{0}=5, \;\ a_{1}=16

We can do subs and find them by performing each iteration. But, there is a way to solve it in general. Then, we just plug in to find whichever number we need in the sequence.

We can change it to a quadratic:

r^{2}-7r+10=0

This factors:

(r-2)(r-5)=0

The roots are r=2 and 5

So, we write a_{n}=b\cdot 2^{n}+c\cdot 5^{n}

Use the initial conditions to find b and c:

a_{0}=5\Rightarrow b\cdot 2^{0}+c\cdot 5^{0}=5\Rightarrow b+c=5

a_{1}=16\Rightarrow b\cdot 2^{1}+c\cdot 5^{1}=16\Rightarrow 2b+5c=16

We solve the little system:

b+c=5

2b+5c=16

b=3, \;\ c=2

a_{n}=3\cdot 2^{n}+2\cdot 5^{n}

There is the formula. If we want the 10th number for instance, we plug in n=10

3\cdot 2^{10}+2\cdot 5^{10}=19,534,322

Just a small sequence tutorial if you like.

Strat, see if you can pick up this sequence

2, 4, 12, 24, 72, 144

432
I see it is Times 2, times 3, times 2, times 3...
I think galactus answered my question, Thanks All.

So, you see the pattern then?

The OP was a sequence much the same. We were just able to see the pattern.

It wasn't I didn't see the solution, I was curious if their was some other way(math wise)