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parallaxicality
2008-Oct-13, 09:07 PM
I've been mulling for years over a numerical sequence that was just namechecked on a history of math I've been watching on TV. Apparently it was first discovered by Muhammad ibn Mūsā al-Khwārizmī, which at least puts me in good company. But I have no idea what it's called or how to generalise it.

Basically, it works like this:

If you take a number and square it, the square will always be one more than the multiple of the two numbers on either side of the square. In other words, if you take the numbers 2, 3, and 4, 3 squared (9) is one more than 2 x 4 (8). If you take the numbers 20, 21, and 22, 21 squared (441) is one more than 20 x 22 (440) and so on.

But then, if you take a sequence of numbers separated by 2, say, 4, 6, and 8, the middle square, 36, is 4 more than 4 x 8 (32). If you take three numbers separated by 3, say 6, 9, and 12, the middle square (81) is nine more than the multiple of the other two numbers (6 x 12 = 72).

So. In any sequence of three numbers separated by a common value, the multiples of the first and third numbers will always be less than the middle number squared by a value equal to the square of the number separating them.

If that makes any sense.

But I was wondering, can this principle be extended? What happens when you move to more complex sequences, such as 2, 4, 8? Or geometric sequences, like 4, 16, 64? Can this sequence be extended beyond three numbers? Is there an ultimate equation that explains the pattern completely?

agingjb
2008-Oct-13, 09:12 PM
(x-1)(x+1) =x^2-x+x-1=x^2-1

(x-a)(x+a) = x^2-ax+ax-a^2=x^2-a^2

("^" represents "squared"; I'm not sure if I can put superscripts here.)

Tim Thompson
2008-Oct-13, 09:21 PM
If you take a number and square it, the square will always be one more than the multiple of the two numbers on either side of the square.
That's easy enough. Three numbers n-1, n, n+1 in sequence. (n-1)*(n+1) = n2 -n +n -1 = n2-1.

Now generalize: 3 numbers in sequence n-a, n, n+a. Now (n-a)*(n+a) = n2-a2 which reproduces all of the other cases you show.

Aristocrates
2008-Oct-13, 09:40 PM
I think Parallaxicality is asking about generalizations to sequences of number separated by means other than adding a. Example:
x-a, x, x+a is what we've been talking about
x(a-1), x(a), x(a+1) would be another sequence
x^(a), x^(2a), x^(3a) would be another

I don't think there's a generalization possible for that, though.

AndreasJ
2008-Oct-13, 09:51 PM
(x-1)(x+1) =x^2-x+x-1=x^2-1

(x-a)(x+a) = x^2-ax+ax-a^2=x^2-a^2

("^" represents "squared"; I'm not sure if I can put superscripts here.)

You can:

(x-a)(x+a) = x2-ax+ax-a2=x2-a2

Simply put around what you want to superscript. :)

a1call
2008-Oct-13, 10:07 PM
I think one generalization is possible:

*- Any odd number of consecutive numbers will have a constant difference between the middle number squared and the last-first multiplied.

For example any 5 consecutive numbers say 2 to 6 hold true for

4^2=16= (2x6)+4

If I'm not mistaking the +4 is always constant for any 5 sequential numbers.

There should also exist relationships for more complex sequences but it would be of the parametrized (variables) form which could be formulated related to one of the numbers in the sequence(say the middle one).

A more general formula would probably require incorporating series, sums and factorials.

agingjb
2008-Oct-14, 08:00 AM
You can:

(x-a)(x+a) = x2-ax+ax-a2=x2-a2

Simply put around what you want to superscript. :)

Thanks2.

mugaliens
2008-Oct-14, 02:39 PM
In any sequence of three numbers separated by a common value, the multiples of the first and third numbers will always be less than the middle number squared by a value equal to the square of the number separating them.

If that makes any sense.

Certainly. Although I'd say it as, "In any sequence of three evenly-spaced numbers, the difference between the square of the middle number and the product of the first and last numbers is equal to the square of the space between each number."

The numbers themselves do not matter. It could be 3, 6, and 9, or 5, 8, and 11. The space between them is 3, and the difference between the center, squared, and the product of the first and last is 9.

Thus, we can state:

Given four rational numbers, a, b, c, and x, such that c=b+x=a+2x, the following is true: b2=a*c+x2

But I was wondering, can this principle be extended? What happens when you move to more complex sequences, such as 2, 4, 8? Or geometric sequences, like 4, 16, 64? Can this sequence be extended beyond three numbers? Is there an ultimate equation that explains the pattern completely?

Off the top of my head, I'd say "no," as the relationship between the space between the three digits and the difference as calculated above is dependant on how that space is calculated.

For a simple progression, such as 2, 4, 6, or 3, 6, 9, the above equation holds. For more complex progressions, such that each succeeding number is a factor or power of the either the previous, or the previous raised to some factor, the equations would be particular to the specific relationship, and would not, therefore, be able to be represented by a single, overarchng equation.

hhEb09'1
2008-Oct-17, 10:52 AM
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But I was wondering, can this principle be extended? What happens when you move to more complex sequences, such as 2, 4, 8? Or geometric sequences, like 4, 16, 64? Can this sequence be extended beyond three numbers? Is there an ultimate equation that explains the pattern completely?Yes, the principle can be extended, especially to the examples you gave there (2, 4, 8 is also geometric, BTW). Unfortunately, the name of the ultimate equation is just "algebra".

Both of your sequences have the algebraic form n, n2, n3, so of course in any similar sequence the square of the middle one is always going to be exactly equal to the product of the other two! We can show that by just performing the operations on the algebraic forms, like Tim Thompson did with your other sequences.
I think one generalization is possible:

*- Any odd number of consecutive numbers will have a constant difference between the middle number squared and the last-first multiplied.

For example any 5 consecutive numbers say 2 to 6 hold true for

4^2=16= (2x6)+4

If I'm not mistaking the +4 is always constant for any 5 sequential numbers.Yahbut you're just ignoring all of the numbers but the middle one and the two on the outside, so those just boil down to the ones that Tim Thompson generalized.