Conjecture regarding perfect numbers.

[MathematicalPhysicist]

From taking breaks from preparing for a talk I have in geometry, I started toying a little bit with perfect numbers.

We all know that 3^3+4^3+5^3=216=6^3

This and the well known pythogrean triplet 3^2+4^2=5^2.

So I thought of toying a little bit with powers of three and two, and I found by coincidence that:
3^3+5^3+7^3=495= 496 -1 , where we all know that 496 is a perfect number.

Then I thought ridicuosly that this can happen also for other perfect numbers, but to no a veil, for 6 we can't have powers of three which are distinct from each other, but we do have powers of 2, 6-1=2^2+1^2.

So I thought to myself, maybe every perfect number minus 1 can be represented as powers of 3 or 2 of distinct natural numbers.

For 28 we have 28-1=3^3=27.

For 8128 we have 8128-1=19^3+8^3+7^3+6^3+5^3+4^3+2^3.

All the above is sheer luck and coincidence, but this raises the conjecture:

Every (even) perfect number minus one can be represented as a sum of distinct powers of 2 or 3

I don't have enough time to check for the next perfect number.

I did the last calculation via google, check me that I don't have mistakes.
Is this already known?

 

[DonAntonio]

From taking breaks from preparing for a talk I have in geometry, I started toying a little bit with perfect numbers.

We all know that 3^3+4^3+5^3=216=6^3

This and the well known pythogrean triplet 3^2+4^2=5^2.

So I thought of toying a little bit with powers of three and two, and I found by coincidence that:
3^3+5^3+7^3=495= 496 -1 , where we all know that 496 is a perfect number.

Then I thought ridicuosly that this can happen also for other perfect numbers, but to no a veil, for 6 we can't have powers of three which are distinct from each other, but we do have powers of 2, 6-1=2^2+1^2.

So I thought to myself, maybe every perfect number minus 1 can be represented as powers of 3 or 2 of distinct natural numbers.

For 28 we have 28-1=3^3=27.

For 8128 we have 8128-1=19^3+8^3+7^3+6^3+5^3+4^3+2^3.

All the above is sheer luck and coincidence, but this raises the conjecture:

Every (even) perfect number minus one can be represented as a sum of distinct powers of 2 or 3

I don't have enough time to check for the next perfect number.

I did the last calculation via google, check me that I don't have mistakes.
Is this already known?

If you don't have enough time to check your own conjecture, and verify it is a

sound one, what makes you think others will invest the needed time?

I, for one, don't know about any results in this direction.

DonAntonio

 

[MathematicalPhysicist]

No need to be angry, Don.

I just seen something at the first four perfect numbers, the next perfect number is of 8 digits.

I don't know how to even start guessing for such a number such a representation.

Anyway, what makes a conjecture a sound one? There aren't a lot of perfect numbers known to us, and those that are huge.

I don't know how to even start programming a code that checks for such a representation.

So I posed this conjecture in hope that someone who is better than me in coding will test this for the next perfect number and so on.
Edit:
Oh, wait I think I have an idea of how to check this in a code, not sure, if anyone can help me on this that would be superb.

I start with a perfect number, subtract from it the number 1, and then I try to use here some loop to subtract from this number distinct powers of 2 or distinct powers of 3 until I come to the number zero in which case I am done, and then the programme should print the powers. Any programmer in the audience, I am a bit rusty with this.

 

[acabus]

These are the numbers that are the sums of distinct squares:
https://oeis.org/A003995
These are the numbers that are not:
https://oeis.org/A001422

And these that are the sums of distinct cubes:
https://oeis.org/A003997
These are the numbers that are not:
https://oeis.org/A001476

From manually checking (so there might be a mistake or two), the only numbers that are in neither the sum of distinct squares or the sum of distinct cubes are:

2,3,6,7,11,15,18,19,22,23,24,31,32,33,43,44,47,48,60,67,76,96,108,112,128.

Making your conjecture true, but not very impressive =]

 

[MathematicalPhysicist]

Thanks, abacus.

I guess there's a long long way until I'll have something new and meaningful to say in mathematics.

 

[Dadface]

Hello Mathematical Physicist.May I recommend the following book which I guess you will find interesting:

THE PENGUIN DICTIONARY OF CURIOUS AND INTERESTING NUMBERS by DAVID WELLS

 

[DonAntonio]

No need to be angry, Don.

{}

Uh? Why would I be angry? I don't get it...

I just seen something at the first four perfect numbers, the next perfect number is of 8 digits.

I don't know how to even start guessing for such a number such a representation.

Anyway, what makes a conjecture a sound one? There aren't a lot of perfect numbers known to us, and those that are huge.

Well, there are about 47 known perfect numbers, which you can check here http://en.wikipedia.org/wiki/List_of_perfect_numbers

A sound conjecture about them would imply to check the conjecture for them all, say. I'm not saying it's easy, I'm not

saying it's cheap or fast, but that's the way things are. If you've access to some computing resources in some university or

stuff then you might get some help there.

I don't know how to even start programming a code that checks for such a representation.

So I posed this conjecture in hope that someone who is better than me in coding will test this for the next perfect number and so on.
Edit:
Oh, wait I think I have an idea of how to check this in a code, not sure, if anyone can help me on this that would be superb.

I start with a perfect number, subtract from it the number 1, and then I try to use here some loop to subtract from this number distinct powers of 2 or distinct powers of 3 until I come to the number zero in which case I am done, and then the programme should print the powers. Any programmer in the audience, I am a bit rusty with this.

That's an idea. Good luck and try it, as it seems doable even with not a a huge computer.

DonAntonio

 

[Anti-Crackpot]

All the above is sheer luck and coincidence, but this raises the conjecture:

Every (even) perfect number minus one can be represented as a sum of distinct powers of 2 or 3

In the spirit of trivial, but true...

Any odd 1-Perfect Number or any even 2-Perfect Number is expressible as the difference between 2 distinct powers of 2.

e.g. 1 = 2- 1, 6 = 8 - 2, 28 = 32 - 4, 496 = 512 - 16, 8128 = 8192 - 64

 

[Eval]

You have piqued my curiosity. I must go to meet a friend, but I will be distracted by trying to figure some of this stuff out. What I can tell you with certainty is that all perfect numbers greater than 6 can be written as the difference of two squares and the same holds for all perfect numbers minus 1. I will see if I can formulate anything from there. (If you would like proof of those claims, I can supply them).

 

[Eval]

The only progress I have had is not much. I can say that in order for the perfect number to be a sum of squares, if there are any odd squares involved, there must be multiples of 4.(because all perfect numbers >6 are divisible by 4 and all odd squares are of the form 4n+1). I have a few other identities formed, but none of them have led me to a proof.

 

[Anti-Crackpot]

FWIW, all Mersenne Prime Exponents > 3 are congruent to 1, 5, 7, 11, 13, 17, 19 or 23 (mod 24), which happens to be the automorphism group for 24 (8 total self-mappings). This follows from the fact that all Mersenne Prime exponents are prime and therefore follow the form n = 6x +/- 1 for n >3.

- AC