# B An odd integer series formula?

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1. May 4, 2016

### Dennis Plews

A few months ago I posted a simple equation that shows an interesting nexus between the difference between the squares of successive integers and the sums of their roots, viz:

Where y = x+1 then (x + y) = (y2 - x2)

Recently I expanded this relationship as follows:

Where n is any integer and y = (x + n), then n(x+y) = (y2 - x2)

Starting with x = 1 and y = 2 and increasing the x and y values by 1 at each iteration, this seems to produce an odd integers sequence as follows:

(1 + 2) = 3 = (4 - 1)
(2 + 3) = 5 = (9 - 4)
(3 + 4) = 7 = (16 - 9)
(4 + 5) = 9 = (25 - 16)
(5 + 6) = 11 = (36 - 25)....

Using the y = (x + n) form with the x value at 1 and increasing y by (x + n) gives a similar result:

1(1 + 2) = 3 = (4 - 1)
2(1 + 3) = 8 = (9 - 1)
3(1 + 4) = 15 = (16 - 1)
4(1 + 5) = 24 = (25 - 1)
5(1 + 6) = 35 = (36 -14)...

The difference between the successive results values being a sequence of odd integers.

Using the y = (x + n) form with the x value at 2 gives a similar result:

1(2 + 3) = 5 = (9 - 4)
2(2 + 4) = 12 = (16 - 4)
3(2 + 5) = 21 = (25 - 4)
4(2 + 6) = 32 = (36 - 4)
5(2 + 7) = 45 = (49 - 4)...

The difference between the successive results values again being a sequence of odd integers.

Not being very sophisticated mathematically I looked through Wikipedia’s integer series page (https://en.wikipedia.org/wiki/Integer_sequence) and found nothing like this series. Fibonacci numbers seem similar. I am curious to learn if this relationship is already known and whether it has any relationship to other known mathematical relationships.

2. May 4, 2016

### Fightfish

To be frank here, I'm not quite sure what the interesting thing about the relationships that you found is - so if I missed it, please do point it out.

Well, if you increased x and y by 1 each, the sum (x+y) increases by 2 for each step, so if u started with an odd value, you will naturally generate a sequence of odd integers.

Well the difference between successive steps is
$[(y+1)^2 - x^2] - [y^2 - x^2] = 2y + 1$
which naturally forms a sequence of successive odd integers since you're increasing $y$ by 1 each step.

3. May 4, 2016

### TeethWhitener

Difference of squares: $y^2-x^2=(y+x)(y-x)$. Substitute $y=x+n$ to get $y^2-x^2=(y+x)n$.

4. May 4, 2016

### SammyS

Staff Emeritus
Yes, a little over nine months ago you started such a thread.

In that thread you discussed the equation x + y = y2 - x2 .

I assume you intend the 2 to be used as an exponent here also.

The equation $\ y = x+n \$ is equivalent to the equation $\ y - x=n \$.

Multiplying that $\ y+x\$ gives $\ (y - x)(y+x) =n(y+x) \$.

This equation has a set of solutions in addition to those for the initial equation. These are the solutions to $\ y=x\$.

Of course, $\ (y - x)(y+x) =n(y+x) \$ is the same as $\ y^2 - x^2 =n(y+x) \$.

5. May 4, 2016

### SammyS

Staff Emeritus
Let f(x) = x2 .

What you have is variations on f(x+1) - f(x), for integer values of x.

For the case with n, it's essentially a similar difference with a cubic function.

6. May 6, 2016

### Dennis Plews

I appreciate all of your comments. I have learned something from each and am encouraged to further my math skills.