# Convert to formula

1. Aug 1, 2010

### smslca

1. The problem statement, all variables and given/known data
please convert the below(P) into formula using (m), or give an algorithm using (m) to get (P)

m=2, P=0
m=3, P=3
m=4, P=3+5
m=5, P=3+5+5
m=6, P=3+5+5+7
m=7, P=3+5+5+7+9
m=8, P=3+5+5+7+9+9
m=9, P=3+5+5+7+9+9+11
m=10, P=3+5+5+7+9+9+11+13
m=11, P=3+5+5+7+9+9+11+13+13
m=12, P=3+5+5+7+9+9+11+13+13+15
m=13, P=3+5+5+7+9+9+11+13+13+15+17
m=14, P=3+5+5+7+9+9+11+13+13+15+17+17
and so on

2. Relevant equations

3. The attempt at a solution
I did not understand how to evaluate it into a formula

Last edited: Aug 2, 2010
2. Aug 2, 2010

### Mentallic

Using induction you can show that

$$3+5+7+...+(2n+1)=n(n+2)$$

and

$$5+9+13+...+(4n+1)=n(2n+3)$$

If you need to prove these yourself then you can manipulate the first sum by turning it into the well known
$$1+3+5+...=n^2$$
and if you need to prove this too... well... use geometry I suppose.

and for the second,
$$=4+8+12+...+4n-n$$
$$=4(1+2+3+...+n)-n$$

Now I couldn't quite find how to merge both formulae together to have a single function P=f(m) for all values of m>1. I'll keep working on it.

3. Aug 2, 2010

### smslca

yeah thanks, I got some view on the series, But I too cannot understand the mixed series. So please keep trying. I also will try on it.

4. Aug 2, 2010

### Mentallic

Ok I figured out a nice pattern and after much blood, sweat and tears, I finally solved it. There will be 3 cases to consider.

Are you familiar with modulo arithmetic?
For m(mod 3)=0, that is, m=3,6,9,12...

We have the pattern:

m=3 , P = (3) + (0)
m=6 , P = (3+5+7) + (5)
m=9 , P = (3+5+7+9+11) + (5+9)
m=12, P = (3+5+7+9+11+13+15) + (5+9+13)

etc.

Now since in this case we don't need to consider what is happening in the other cases and we are dealing with m that are multiples of 3, let's take $$\frac{m}{3}=x$$. So for m=12, x=4 and you can consider that as being the 4th sequence in this lone pattern.

for x=1, we have 1 in the first bracket, and 0 in the second
for x=2, we have 3 in the first bracket, and 1 in the second
for x=3, we have 5 in the first bracket, and 2 in the second
for x=2, we have 7 in the first bracket, and 3 in the second

Can you see the pattern arising?

for x=x, we have (2x-1) in the first bracket, and (x-1) in the second

Now looking at the formulae I gave you earlier,

$$3+5+7+...+(2n+1)=n(n+2)$$
and

$$5+9+13+...+(4n+1)=n(2n+3)$$

Let the first bracket with the first formula be n=2x-1. What this means is we have to substitute n=2x-1 into the sum formula n(n-2).
Secondly, substitute n=x-1 in the second formula.

So we obtain $$P=(2x-1)(2x+1)+(x-1)(2x+1)=(3x-2)(2x+1)$$

That's the formula for m(mod 3)=0 where m/3=x.

So let's take m=12, thus x=4.

m=12, P=3+5+5+7+9+9+11+13+13+15=90

P=(3.4-2)(2.4+1)=10.9=90

It works for all others too

Now maybe you can try apply this idea to the other two cases? Let m(mod 3)=1, i.e. m=4,7,10... and $$\frac{m-1}{3}=x$$.

For m(mod 3)=2 let $$\frac{m+1}{3}=x$$

This will finally satisfy all your summations.

5. Aug 2, 2010

### smslca

thanks for the replies

6. Aug 3, 2010

### Mentallic

No problem. Are you able to find the answer to the other two cases?

7. Aug 3, 2010

### smslca

yeah I got it from other forum.as

Let n = [m/3] ; [] mean integer part
and k = m mod 3 ; mod mean modulus (remainder of the integer divide m/3)

That is:
m --> n , k
3 --> 1, 0
4 --> 1, 1
5 --> 1, 2
6 --> 2, 0
7 --> 2, 1
8 --> 2, 2
9 --> 3, 0
...............

Then P(n,k) = n*(6*n-1) + k*(4*n+1) - 2