Number theory conjecture... :D

Question

empty · Answer

$\tau(n)$ is the divisor counting function, for example $\tau(12)=6$ since there are six divisors of 12: 1, 2, 3, 4, 6, 12.

Conjecture: There exists one and only one number $n$ with an odd number of divisors $\tau(n) \equiv 1 \mod 2$ such that its total number of divisors is equal to the average of the adjacent divisors.

$$\tau(n)=\frac{\tau(n+1)+\tau(n-1)}{2}$$

That number is $n=8836$ with $\tau(8836)=9$ which is odd.

So, proof or counter example, any ideas? :D

usukidoll · Answer

I feel like putting that n = 8836 just to verify that it's correct but it's a shoddish way to prove it.. 
because I was thinking that we have n = 12
and then into that formula I would need to find tao(13) and tao(11) which don't have that many divisors and both of those numbers 13 and 11 are prime.

empty · Answer

I tested all numbers up to 100,000 and this is the only number that satisfies this. I could go higher I guess haha.

usukidoll · Answer

$$\tau(12)=\frac{\tau(13)+\tau(11)}{2}$$ but then for 13 only 1 and 13 
and 11 only 1 and 11 
that's like 2 a piece. wtheck I got 2.

empty · Answer

Wait a sec, that's not right, 

$$\tau(11)=2$$$$\tau(12)=6$$$$\tau(13)=2$$
$$6 \ne \frac{2+2}{2}$$

Although you did hit up on something interesting hmm.

empty · Answer

I didn't really think about twin primes, they seem to be kind of related to this.

usukidoll · Answer

yeah I know that 
$$6 \ne \frac{2+2}{2} $$

but 13 and 11 are both prime numbers . the only numbers divisible are 1 and itself. 

so for 11
1 and 11
so for 13
1 and 13

empty · Answer

Oh I guess I'm not following why you wrote this: $$\tau(12)=\frac{\tau(13)+\tau(11)}{2}$$ 

Also I just realized, in order for 

$$\tau(n) \equiv 1 \mod 2$$ then n has to be a perfect square. That cuts out a ton of cases haha.

usukidoll · Answer

damn bro you didn't mention that. so n has to be perfect square numbers
4 25 36 la la la

empty · Answer

Well it doesn't HAVE to be, I just figured this out! hahaha

empty · Answer

Like look at 4: 1, 2, 4. The only way you can get an odd number of divisors is if the number is a perfect square, so it's the same condition really.

empty · Answer

Same exact conjecture, just said more simply:

Conjecture: There is only one perfect square such that its total number of divisors is equal to the average of the adjacent numbers divisors.

usukidoll · Answer

oh O_O!

usukidoll · Answer

25: 1 5 25 
36: 1 6 36 (but for this number I swear the list is longer)

empty · Answer

haha yeah:

1, 2, 3, 4, 6, 9, 36

usukidoll · Answer

49 might work
49: 1 7 49

empty · Answer

Ok so there's a rick to evaluating these things,

$$\tau(ab)=\tau(a)\tau(b)$$ if $\gcd(a,b)=1$.

So an example of this is probably much easier, and basically all this means is you can break it up into each of the prime factors like this:

$$\tau(12)=\tau(4)*\tau(3) = 3*2 =6$$

More generally but less general than the original first thing I put:

$$\tau(p^xq^y) = \tau(p^x)\tau(q^y) = (x+1)(y+1)$$

in fact for a prime, 

$$\tau(p^k) = k+1$$

empty · Answer

OK another alternate notation is $\tau(n) = \sigma_0(n)$.... why do I say that? Cause check this out: http://www.wolframalpha.com/input/?t=crmtb01&i=tau(1468%5E2)+%3D+(+tau(1468%5E2%2B1)%2Btau(1468%5E2-1))%2F2