Some time ago i requested a math forum but there did not seem to be enough interest. Recently i viewed some posts on other threads that supported this idea.

I am not now requesting the same thing but i wish to ask a question.

Question:

Can anyone one this board tell me the next number in this sequence?

{1,2,2,4,2........
Posted on 2002-08-21 22:06:34 by IwasTitan
There are too few numbers there to know exactly which the next number is. Here are some of my guesses:

6: (1),2,(2),4,(2),[6]
2: if the sequence is symmetric: 1,2,2,4,2,[2,1]
4: if the even numbers are twice as large as the previous number
Posted on 2002-08-22 01:03:12 by gliptic
4
Posted on 2002-08-22 01:51:23 by SpooK

Can anyone one this board tell me the next number in this sequence?

{1,2,2,4,2........

Looks like your are playing with primes :)

Primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31... etc.
Correct sequence: 1, 2, 2, 4, 2, 4, 2, 4, 6, 2... etc.

It's the difference between a prime and the next one.
Posted on 2002-08-22 02:53:39 by bazik
As fonzy would say ..."Correctamondo"

I wonder if the sequence is continued all numbers are even.

:alright:
Posted on 2002-08-22 11:17:23 by IwasTitan
All prime numbers, with the exception of 2, are by their definition odd.
All even numbers are divisable by 2, and so cannot be prime (except 2).

Any odd number can be written as (2*n -1) where n is some integer.
So we have (2*X - 1) - (2*Y - 1) = 2X - 2Y which must be even!

Thus we know that the difference between any two primes (as long as neither of the two primes involved is 2) must be an even number.

Mirno
Posted on 2002-08-22 11:23:55 by Mirno
Mirno,
nice explanation! :)
Posted on 2002-08-22 11:49:17 by bazik
Yes Mirno that was very nicely done. Putting it into concrete formula that is. Maybe you could take a shot at this one.

Take as a sequence the difference between the squares of a prime and the previouse one.

(e.g. the first number in the sequence is defined as 3^2 - 2^2 = 9-4=5)

The sequence becomes:

{5,16,24,72,48,120,72,168,312,120,408,312,168,360,600,672,240,248,1072,288,912,648,1032,1488,792,408}

Notice that some numbers repeat.The first to repeat is 72. The next are 120,168,312,408.

Now take the digits of these numbers and add then together. If the result is greater then 10 then continue the process.

72 = 7 + 2 = 9
120 = 1 + 2 + 0 =3
168 = 1 + 6 + 8 = 15 = 1 + 5 = 6
312 = 3 + 1 + 2 = 6
408 = 4 + 0 + 8 = 12 = 1 + 2 =3

All multiples of three.

In the sequence many of the numbers can be reduced to multiples of three but some can not, namely the first two and 1072. The question is whether all the numbers that do repeat reduce in this fashion to a multiple of 3. Also it is obviose that if the digits of a repeated value reduced to 10 when added then this would blow the theory into the trash basket.(i.e. If 1072 repeated.

Mirno do you have formula up your sleeve to prove or disprove this? For myself it would be interesting to know.

:alright:
Posted on 2002-08-22 12:58:17 by IwasTitan
Besides the games, are you intending to discuss Algebra, Clacululs and etc. in that new Forum?
I would like to see a Math Forum here, too.
Posted on 2002-08-23 16:32:25 by wolfao
Who is that girl on your avatar, wolfao?
Posted on 2002-08-23 17:50:37 by comrade

Who is that girl on your avatar, wolfao?

Posted on 2002-08-23 18:03:38 by bazik
Danke :grin:
Posted on 2002-08-23 19:08:24 by comrade
wolfao...yes i would like to discuss those things if there were a math forum.

And in a calculas discussion i would post this:

(x^2 +2)^2.dx

(x +2)^100.dx

Why is the second so easy to integrate whereas the first has to be expanded. Expanding the second creates an equation far more complex and of much higher power than the first yet because the X in the brackets is only of power 1 we can use a simple formula and the answer is:

1/101*(x+2)^101 + C

There is presently no method like this for the first equation because x is to the power of 2 even though it is less complex when expanded. I'm beginning to think that these two equations describe numbers that belong to two entirely different sets. Something along the theory of duality.

:alright:
Posted on 2002-08-23 19:21:49 by IwasTitan
It is:

1/3 (x^2+2)^3 / 2x.....

there is a simple rule!

Try diff you get... (x^2_2)^2.
Posted on 2002-08-26 09:25:15 by mega
mega...sorry but i don't get yah..by your method your solution would be ok for this integral:

(x^2+2)^1/2x

but not for the one i posted.

(x^2+2)^2

You have to subtract 1 from the power outside the brackets when you differentiate. Your value is 3/2. 3/2 - 1 = 1/2 which is not 2 which is what is in the original equation that we want to integrate. Sorry there is no simple solution that i know of at this time to integrate without expansion.

Also i'm not really sure that your manner of using 3/2x as your power complies with the rules of integration and differentiation. What you have done is subracted 1 from the constant in the power...not 1 from the power itself ( 3/2x -1). I have never seen it used in that manner and i doubt that it is a legal operation in differentiation. Feel free to enlighten me though.

Also your equation defines an infinite set of equations because every time the value x changes your equation changes. When x is 2 your power is 3. When x is 4 your power is 6 etc. This a long way from finding the rate of change or the area under a curve for a single equation.

:alright:
Posted on 2002-08-26 13:55:23 by IwasTitan
The second one can be integrated so simply because it's a special case of some general formula (I will use "I" letter for the integration symbol):

I f^n df = /(n+1)

Well, it's generally used with x instead of f, but I wrote it this way, because now I'm going to use f(x) instead of x, df(x) is - of course - the same as f'(x)dx, so the rewritten formula is:

I f'(x)dx = /(n+1)

And your second is the special case

I [(x+2)^100]1dx = [(x+2)^101]/101

Your first one should be multiplied by 2x to be solvable this way.

Playing with elementary differentation/integration is really fun, I wish I had to play only with such simple math today...
Posted on 2002-08-26 14:18:03 by Tomasz Grysztar
Your first one should be multiplied by 2x to be solvable this way.

Privalov show me what you mean by this. I understand your equations but i still don't see a simple solution for integrating the first. (x^2+2)^2 . If i multiply it by 2x before integrating then it is not the same equation any more.

oh i see you multiply it by the reciprocal of 2x after the integration.

(X^2 +2)^2.dx = 1/3(X^2+2)^3 * 1/2x = 1/6x(X^2+2)^3 +c

dy/dx 1/6x(X^2+2)^3 = 6x/6x(X^2+2)^2 = (X^2+2)^2

Yes know i remember.

So just what were those ashol*s talking about in my calculas course so many years ago. They lie to pick peoples brains.

:alright

:alright:
Posted on 2002-08-26 14:32:07 by IwasTitan
Yes, after multiplying by 2x it's not the same problem - that's what I meant, because you asked what differs these two, that the second is simple to solve, and the second one isn't. That's because second one fits in that general rule, and the first doesn't.
Posted on 2002-08-26 14:39:03 by Tomasz Grysztar
Privalov...so what i posted in my last edit of the previouse post as being the solution really isn't ?

This is confusing.

:alright:
Posted on 2002-08-26 14:48:54 by IwasTitan
i posted the correct answer... just wrote it confusingly

1/3 (x^2+2)^3 / 2x.....

is really

(x^2+2)^3
-------------
6x

or 1/3 ( (x^2+2)^3 ) / 2x.....

The general formula is

integrate ( f(x) )^n

= ( f(x) )^n+1
------------------
f'(x) * (n+1)

is only we could write Latex...
Posted on 2002-08-26 16:56:01 by mega