# Generating Odd Numbers

• Nov 24th 2010, 05:09 AM
Totype
Generating Odd Numbers
x and x2 are both odd numbers in the formula
1.5x + .5 = x2
The purpose of this formula is to generate and odd number x2
My question is whether it is possible to generate odd numbers every time this formula is run (with x2 becoming x the next time its run)

so far this is what I've done

x2 can be expressed as 2n+1 where n is an integer.

1.5x + .5 = 2n+1
1.5x = 2n +.5
3x = 4n + 1
x = n + n/3 + 1/3

The remainder of n/3 must 2/3 otherwise x will not be an integer

n must be a number which is a multiple of 3 with 2 added to it
which can be expressed as 3k + 2

x = 4*(3k+2)/3 + 1/3
x = 4k + 8/3 + 1/3
x = 4k + 3

so when x is a multiple of 4 with 3 added to it an odd number will be produced

Since I want to keep on producing odd numbers, I then checked what values of x produce a number in the form of 4j+3

1.5x + .5 = x2
1.5x + .5 = 4j+3
1.5(4k+3)+.5 = 4j+3
6k + 5 = 4j+3
6k + 2 = 4j
1.5k + .5 = j
.5k must produce a remainder of .5 if j is to be an integer, therefore k must be an odd number

To recap, the formula 1.5x + .5 = x2 produces an odd number when x is written in the form 4k+3,
and if I want to produce another odd number from when this formula is applied to x2 then k must be an odd number.

So now to check what sort of odd numbers produce x2 in the form 4j+3 so that j is an odd number, we'll express k as 2n+1 and j as 2m + 1

1.5x + .5 = x2
1.5(4j+3)+.5 = 4k + 3
1.5(4(2n+1)+3)+.5 = 4(2m+1)+3
1.5(8n + 4 + 3)+.5 = 8m + 4 + 3
1.5(8n + 7)+.5 = 8m + 4 + 3
12n + 10.5 + .5 = 8m+ 7
12n + 11 = 8m + 7
12n + 4 = 8m
1.5n + ,5 = m
half of of n must be equal to .5 for m to be an integer

so n must be an odd number

for m to be an odd number then .5n + .5 must be an odd number since n is even and odd + even = odd
.5n + .5 = 2h
n = 4h - 1
so n must be a number that is a multiple of 4 with 1 taken away from it, or a multiple of 4 with 3 added to it....

Now I'm stuck, any help? Anyone know an easier way? It just seems to be getting deeper and deeper in a circle...
I'm presuming its impossible since the number x can only be complicated up to a certain extent so after a certain
point the number generated by the formula will be even

Is this right?
• Nov 24th 2010, 05:50 AM
emakarov
Sorry, have not read everything. If you need a recurrence relation for generating all odd numbers, you can use

$x_1=1$
$x_{n+1}=x_n+2$
• Nov 24th 2010, 04:50 PM
Totype
Thanks for the link its pretty useful. Though I got this equation from a broader problem where I need to check whether this particular equation will generate a series of odd numbers (not necessarily all odd numbers) before I can move on.
• Nov 25th 2010, 12:34 PM
emakarov
OK, I also get that $x_{n+1}$ is odd iff $x_n\equiv3\pmod{4}$, i.e., $x_n$ leaves the remainder 3 when divided by 4. Similarly, $x_{n+2}$ is odd iff $x_n\equiv7\pmod{8}$. In general, $x_{n+k}$ is odd iff $x_n\equiv2^{k+1}-1\pmod{2^{k+1}}$.

This means that for every $x_0$ there exists an $n$ such that $x_n$ is even (and $x_{n+1}$ is not an integer). Indeed, let $n$ be such that $x_0<2^n$. Then $x_0\not\equiv 2^{n+1}-1\pmod{2^{n+1}}$, i.e., $x_n$ is not odd.