Strange Numbers

Turtle · 09-23-2007

Quote:

Originally Posted by CraigD

I’ll mull it over. I’ll try a dumb arbitrary precision calculator disproof until something smart comes of the mulling – we may get lucky, though a really dumb approach looks to be pretty computationally intense.

Thanks Craig.

I live for computationally intense.

In the mean time, less intense than the odd powers of 2, but worth mentioning, is that all even powers of 2 are Perfect Squares. Is it correct to say "by the rules of exponents" and write 2^2n = 2^n*2^n ?

Qfwfq · 09-24-2007

Certainly, $2^{2n}=(2^n)^2$ and, also, the powers of two which are cubes are all and only those where the exponent is a multiple of 3:

$2^{3n}=(2^n)^3$

Turtle · 09-24-2007

Quote:

Originally Posted by Qfwfq

Certainly, $2^{2n}=(2^n)^2$ and, also, the powers of two which are cubes are all and only those where the exponent is a multiple of 3:

$2^{3n}=(2^n)^3$

Nice Q; thanks. I see in checking my list and every exponent-a-multiple-of-3, that I have some of the cubes labled as the sum of a Perfect Number and a Square; apparently I didn't try taking the cube root.
e.g. 2^33: 8589934592: (1*8589869056)+65536
2^9: 512: Perfect 496 + Square 16: (1*496)+16

As these two are actually cubes ( $512=8^3$ and $8589934592=2048^3$ , I'm compelled to check the other cubes to see if they have a representation as the sum of a Perfect and a Square. Perhaps this is a rule?

What do you think Q of my Conjecture for odd powers of 2?
Conjecture: All odd powers of 2 are either a Cube of an odd power of 2, or the sum of a Perfect-Square-multiple of a Perfect number and a Perfect Square. [(Square*Perfect)+Square]

Turtle · 09-24-2007

Quote:

Originally Posted by Turtle

...
As these two are actually cubes ( $512=8^3$ and $8589934592=2048^3$ , I'm compelled to check the other cubes to see if they have a representation as the sum of a Perfect and a Square. Perhaps this is a rule? ...

Interesting. The Cube 8 has no representation as the sum of a Perfect multiple and a Square. The Cube 2097152 [2^21] has a representation as a sum of a Perfect multiple and a Square: (4228*496)+64, however the multiplier -4228- is not square. Worth noting too that 496 is not the largest Perfect less than 2097152, but 8128 doesn't provide a solution.

The Conjecture is unaffected; please return your seats to the mulling position. The smoking light is on.

CraigD · 09-25-2007

Quote:

Originally Posted by Turtle

[new]Conjecture: All odd powers of 2 are either a Cube of an odd power of 2, or the sum of a Perfect-Square-multiple of a Perfect number and a Perfect Square. [(Square*Perfect)+Square] …

Quote:

Originally Posted by CraigD

I’ll mull it over. I’ll try a dumb arbitrary precision calculator disproof until something smart comes of the mulling – we may get lucky, though a really dumb approach looks to be pretty computationally intense.

It took me more “smarts” to get “computationally intense” down to something manageable enough to produce even a small test. I need to revise my approach, as I’ve hit a computational bottleneck with my counting approach that’s preventing it from getting very high in a reasonable amount of time.

I checked the all $2^{2n+1}$ where $2n+1 \not= 3m$ (the first “or” condition of the conjecture) up to $2^{85}$ , and found no counterexamples. Here are those odd powers of two in terms of

Code:

2^5=28*1^2+2^2
2^7=28*1^2+10^2
2^11=496*2^2+8^2
2^13=8128*1^2+8^2
2^17=8128*4^2+32^2
2^19=8128*8^2+64^2
2^23=8128*31^2+760^2
2^25=33550336*1^2+64^2
2^29=33550336*4^2+256^2
2^31=33550336*8^2+512^2
2^35=8589869056*2^2+512^2
2^37=137438691328*1^2+512^2
2^41=137438691328*4^2+2048^2
2^43=137438691328*8^2+4096^2
2^47=137438691328*32^2+16384^2
2^49=137438691328*64^2+32768^2
2^53=137438691328*256^2+131072^2
2^55=137438691328*512^2+262144^2
2^59=137438691328*2048^2+1048576^2
2^61=2305843008139952128*1^2+32768^2
2^65=2305843008139952128*4^2+131072^2
2^67=2305843008139952128*8^2+262144^2
2^71=2305843008139952128*32^2+1048576^2
2^73=2305843008139952128*64^2+2097152^2
2^77=2305843008139952128*256^2+8388608^2
2^79=2305843008139952128*512^2+16777216^2
2^83=2305843008139952128*2048^2+67108864^2
2^85=2305843008139952128*4096^2+134217728^2

There are lots of coincidences here – such as, with an exception at $2^{23}$ , all the perfect squares being of powers of two – which I super-strongly suspect is due to the known perfect numbers having the form $2^{n-1}(2^n-1)$ . For example, consider:
$2^{85}=2305843008139952128 \cdot 4096^2 +134217728^2$
$2^{85}=2305843008139952128 \cdot (2^{12})^2 +(2^{27})^2$
$2^{85}=(2^{12})^2 \cdot ( 2305843008139952128 +(2^{15})^2)$
$2^{85}=(2^{27})^2 \cdot (2147483647 +1)$
$2^{85}=2^{54} \cdot ((2^{31}-1) +1)$
$2^{85}=2^{54} \cdot 2^{31}$

A proof seems not too far off

Turtle · 09-25-2007

Quote:

Originally Posted by CraigD

It took me more “smarts” to get “computationally intense” down to something manageable enough to produce even a small test. I need to revise my approach, as I’ve hit a computational bottleneck with my counting approach that’s preventing it from getting very high in a reasonable amount of time.

I checked the all $2^{2n+1}$ where $2n+1 not= 3m$ (the first “or” condition of the conjecture) up to $2^{85}$ , and found no counterexamples. Here are those odd powers of two in terms of

Code:

2^5=28*1^2+2^2
2^7=28*1^2+10^2
2^11=496*2^2+8^2
2^13=8128*1^2+8^2
2^17=8128*4^2+32^2
2^19=8128*8^2+64^2
2^23=8128*31^2+760^2
2^25=33550336*1^2+64^2
2^29=33550336*4^2+256^2
2^31=33550336*8^2+512^2
2^35=8589869056*2^2+512^2
2^37=137438691328*1^2+512^2
2^41=137438691328*4^2+2048^2
2^43=137438691328*8^2+4096^2
2^47=137438691328*32^2+16384^2
2^49=137438691328*64^2+32768^2
2^53=137438691328*256^2+131072^2
2^55=137438691328*512^2+262144^2
2^59=137438691328*2048^2+1048576^2
2^61=2305843008139952128*1^2+32768^2
2^65=2305843008139952128*4^2+131072^2
2^67=2305843008139952128*8^2+262144^2
2^71=2305843008139952128*32^2+1048576^2
2^73=2305843008139952128*64^2+2097152^2
2^77=2305843008139952128*256^2+8388608^2
2^79=2305843008139952128*512^2+16777216^2
2^83=2305843008139952128*2048^2+67108864^2
2^85=2305843008139952128*4096^2+134217728^2

There are lots of coincidences here – such as, with an exception at $2^{23}$ , all the perfect squares being of powers of two – which I super-strongly suspect is due to the known perfect numbers having the form $2^{n-1}(2^n-1)$ . For example, consider:
$2^{85}=2305843008139952128 cdot 4096^2 +134217728^2$
$2^{85}=2305843008139952128 cdot (2^{12})^2 +(2^{27})^2$
$2^{85}=(2^{12})^2 cdot ( 2305843008139952128 +(2^{15})^2)$
$2^{85}=(2^{27})^2 cdot (2147483647 +1)$
$2^{85}=2^{54} cdot ((2^{31}-1) +1)$
$2^{85}=2^{54} cdot 2^{31}$

A proof seems not too far off

Looks like my afternoon is now scheduled for comparing lists.

One bit has caught my attention already and I bolded it.

In my list I don't have 2^23 as an exception: I wrote it as (32^2 * 8128) + 256^2 = (1024*8128)+65536. Fascinating...thanks for the counter-mull Craig!

Turtle · 09-25-2007

Quote:

Originally Posted by CraigD

I checked the all $2^{2n+1}$ where $2n+1 not= 3m$ (the first “or” condition of the conjecture) ...

I think it is worth including the Cubes because, except for 2^3=8, they have a representation as (Square*Perfect)+Square as well. (As far as I found)

Quote:

Originally Posted by Craig

2^17=8128*4^2+32^2

In my list I wrote it as 28*1^2+362^2: Inasmuch as 362 isn't square and 28 is not the largest Perfect smaller than 2^17, yours is the correct form.

Quote:

Originally Posted by Craig

A proof seems not too far off

Ahhhh Shangri-La!

If you go back to post #157 you'll find where TimLou took interest and in the posts following he develops his expressions that Alexander has kindly now expanded for our perusal. The Mersenne Prime representation of Perfects is a primary part of Tim's reasoning.

Turtle · 09-25-2007

[old]Conjecture: All odd powers of 2 are either a Cube of an odd power of 2, or the sum of a Perfect-Square-multiple of a Perfect number and a Perfect Square. [(Square*Perfect)+Square]

[new]Conjecture: All odd powers of 2 are the Cube of an odd power of 2, or the sum of a Perfect-Square-multiple of a Perfect number and a Perfect Square [(Square*Perfect)+Square], or both a Cube and Square*Perfect+Square.

This seems a proper amendment. oui/no?

PS I removed the word "either" as we now have 3 choices, and changed the "a" to "the" prefacing "Cube".

CraigD · 09-25-2007

Quote:

Originally Posted by Turtle

[old]Conjecture: All odd powers of 2 are either a Cube of an odd power of 2, or the sum of a Perfect-Square-multiple of a Perfect number and a Perfect Square. [(Square*Perfect)+Square]

The conjecture can be simplified to:
All odd powers of 2 greater than 32 are at least one Perfect-Square-multiple of a Perfect number plus a Perfect Square.

We can prove this as follows:
$2^{2a+1} = 2^{n-1}(2^n-1)2^{2j} +2^{2k}$
We are allowed to select $j$ and $k$ for a given $a$ and $n$ , so may define them as follows:
$k = \frac{n-1 +2j}{2}$
then
$2^{2a+1} = 2^{n-1}(2^n-1)2^{2j} +2^{n-1 +2j}$
$2^{2a+1} = 2^{n-1 +2j}(2^n-1+1)$
$2^{2a+1} = 2^{2n-1 +2j}$
giving
$2a +1 = 2n -1 +2j$
so
$j = a -n +1$
$k = \frac{2a +1 –n}2$
Have positive integer values for any odd value of $n$

According to a proof by Euler, $2^{n-1}(2^n-1)$ is a perfect number if $2^n-1$ is prime (a Mersenne prime). So the conjecture holds for any of the known perfect numbers greater than $6$ ( $2^{2-1}(2^2-1)$ ).

We can use this to quickly generate lists like this

Code:

2^7=(2^2)(2^3-1)((2^1)^2)+(2^2)^2
2^9=(2^2)(2^3-1)((2^2)^2)+(2^3)^2
2^11=(2^4)(2^5-1)((2^1)^2)+(2^3)^2
2^13=(2^4)(2^5-1)((2^2)^2)+(2^4)^2
2^15=(2^6)(2^7-1)((2^1)^2)+(2^4)^2
2^17=(2^6)(2^7-1)((2^2)^2)+(2^5)^2
2^19=(2^6)(2^7-1)((2^3)^2)+(2^6)^2
2^21=(2^6)(2^7-1)((2^4)^2)+(2^7)^2
2^23=(2^6)(2^7-1)((2^5)^2)+(2^8)^2
2^25=(2^6)(2^7-1)((2^6)^2)+(2^9)^2
2^27=(2^12)(2^13-1)((2^1)^2)+(2^7)^2
2^29=(2^12)(2^13-1)((2^2)^2)+(2^8)^2
2^31=(2^12)(2^13-1)((2^3)^2)+(2^9)^2
2^33=(2^12)(2^13-1)((2^4)^2)+(2^10)^2
2^35=(2^16)(2^17-1)((2^1)^2)+(2^9)^2
2^37=(2^16)(2^17-1)((2^2)^2)+(2^10)^2
2^39=(2^18)(2^19-1)((2^1)^2)+(2^10)^2
2^41=(2^18)(2^19-1)((2^2)^2)+(2^11)^2
2^43=(2^18)(2^19-1)((2^3)^2)+(2^12)^2
2^45=(2^18)(2^19-1)((2^4)^2)+(2^13)^2
2^47=(2^18)(2^19-1)((2^5)^2)+(2^14)^2
2^49=(2^18)(2^19-1)((2^6)^2)+(2^15)^2
2^51=(2^18)(2^19-1)((2^7)^2)+(2^16)^2
2^53=(2^18)(2^19-1)((2^8)^2)+(2^17)^2
2^55=(2^18)(2^19-1)((2^9)^2)+(2^18)^2
2^57=(2^18)(2^19-1)((2^10)^2)+(2^19)^2
2^59=(2^18)(2^19-1)((2^11)^2)+(2^20)^2
2^61=(2^18)(2^19-1)((2^12)^2)+(2^21)^2
2^63=(2^30)(2^31-1)((2^1)^2)+(2^16)^2
2^65=(2^30)(2^31-1)((2^2)^2)+(2^17)^2
2^67=(2^30)(2^31-1)((2^3)^2)+(2^18)^2
2^69=(2^30)(2^31-1)((2^4)^2)+(2^19)^2
2^71=(2^30)(2^31-1)((2^5)^2)+(2^20)^2
2^73=(2^30)(2^31-1)((2^6)^2)+(2^21)^2
2^75=(2^30)(2^31-1)((2^7)^2)+(2^22)^2
2^77=(2^30)(2^31-1)((2^8)^2)+(2^23)^2
2^79=(2^30)(2^31-1)((2^9)^2)+(2^24)^2
2^81=(2^30)(2^31-1)((2^10)^2)+(2^25)^2
2^83=(2^30)(2^31-1)((2^11)^2)+(2^26)^2
2^85=(2^30)(2^31-1)((2^12)^2)+(2^27)^2
2^87=(2^30)(2^31-1)((2^13)^2)+(2^28)^2
2^89=(2^30)(2^31-1)((2^14)^2)+(2^29)^2
2^91=(2^30)(2^31-1)((2^15)^2)+(2^30)^2
2^93=(2^30)(2^31-1)((2^16)^2)+(2^31)^2
2^95=(2^30)(2^31-1)((2^17)^2)+(2^32)^2
2^97=(2^30)(2^31-1)((2^18)^2)+(2^33)^2
2^99=(2^30)(2^31-1)((2^19)^2)+(2^34)^2

, or construct some spectacularly large example, such as
$2^{65165315}= 2^{32582657}(2^{32582656}-1) \cdot 2^2 + (2^{8145665})^2$ , a 9808358 digit decimal numeral.

Note that, for values larger than $2^9$ , solutions in this form aren’t unique. For example:

Code:

2^7=(2^2)(2^3-1)((2^1)^2)+(2^2)^2
2^9=(2^2)(2^3-1)((2^2)^2)+(2^3)^2
2^11=(2^4)(2^5-1)((2^1)^2)+(2^3)^2
2^11=(2^2)(2^3-1)((2^3)^2)+(2^4)^2
2^13=(2^4)(2^5-1)((2^2)^2)+(2^4)^2
2^13=(2^2)(2^3-1)((2^4)^2)+(2^5)^2
2^15=(2^6)(2^7-1)((2^1)^2)+(2^4)^2
2^15=(2^4)(2^5-1)((2^3)^2)+(2^5)^2
2^15=(2^2)(2^3-1)((2^5)^2)+(2^6)^2
2^17=(2^6)(2^7-1)((2^2)^2)+(2^5)^2
2^17=(2^4)(2^5-1)((2^4)^2)+(2^6)^2
2^17=(2^2)(2^3-1)((2^6)^2)+(2^7)^2
2^19=(2^6)(2^7-1)((2^3)^2)+(2^6)^2
2^19=(2^4)(2^5-1)((2^5)^2)+(2^7)^2
2^19=(2^2)(2^3-1)((2^7)^2)+(2^8)^2
2^21=(2^6)(2^7-1)((2^4)^2)+(2^7)^2
2^21=(2^4)(2^5-1)((2^6)^2)+(2^8)^2
2^21=(2^2)(2^3-1)((2^8)^2)+(2^9)^2
2^23=(2^6)(2^7-1)((2^5)^2)+(2^8)^2
2^23=(2^4)(2^5-1)((2^7)^2)+(2^9)^2
2^23=(2^2)(2^3-1)((2^9)^2)+(2^10)^2
2^25=(2^6)(2^7-1)((2^6)^2)+(2^9)^2
2^25=(2^4)(2^5-1)((2^8)^2)+(2^10)^2
2^25=(2^2)(2^3-1)((2^10)^2)+(2^11)^2
2^27=(2^12)(2^13-1)((2^1)^2)+(2^7)^2
2^27=(2^6)(2^7-1)((2^7)^2)+(2^10)^2
2^27=(2^4)(2^5-1)((2^9)^2)+(2^11)^2
2^27=(2^2)(2^3-1)((2^11)^2)+(2^12)^2
2^29=(2^12)(2^13-1)((2^2)^2)+(2^8)^2
2^29=(2^6)(2^7-1)((2^8)^2)+(2^11)^2
2^29=(2^4)(2^5-1)((2^10)^2)+(2^12)^2
2^29=(2^2)(2^3-1)((2^12)^2)+(2^13)^2
2^31=(2^12)(2^13-1)((2^3)^2)+(2^9)^2
2^31=(2^6)(2^7-1)((2^9)^2)+(2^12)^2
2^31=(2^4)(2^5-1)((2^11)^2)+(2^13)^2
2^31=(2^2)(2^3-1)((2^13)^2)+(2^14)^2
2^33=(2^12)(2^13-1)((2^4)^2)+(2^10)^2
2^33=(2^6)(2^7-1)((2^10)^2)+(2^13)^2
2^33=(2^4)(2^5-1)((2^12)^2)+(2^14)^2
2^33=(2^2)(2^3-1)((2^14)^2)+(2^15)^2

Turtle · 09-25-2007

Quote:

Originally Posted by CraigD

The conjecture can be simplified to:
All odd powers of 2 greater than 32 are at least one Perfect-Square-multiple of a Perfect number plus a Perfect Square.

We can prove this as follows:
$2^{2a+1} = 2^{n-1}(2^n-1)2^{2j} +2^{2k}$
We are allowed to select $j$ and $k$ for a given $a$ and $n$ , so may define them as follows:
$k = frac{n-1 +2j}{2}$
then
$2^{2a+1} = 2^{n-1}(2^n-1)2^{2j} +2^{n-1 +2j}$
$2^{2a+1} = 2^{n-1 +2j}(2^n-1+1)$
$2^{2a+1} = 2^{2n-1 +2j}$
giving
$2a +1 = 2n -1 +2j$
so
$j = a -n +1$
$k = frac{2a +1 –n}2$
Have positive integer values for any odd value of $n$

According to a proof by Euler, $2^{n-1}(2^n-1)$ is a perfect number if $2^n-1$ is prime (a Mersenne prime). So the conjecture holds for any of the known perfect numbers greater than $6$ ( $2^{2-1}(2^2-1)$ ).

Most excellent. Nuthin' more to do tonight than settle the dust with a cold one. Muchas gracias Craig.

PS looks like the latex doesn't render well in a quote. can we vulcanize it?

Strange Numbers

Hypography?

Share the love!

Sponsors

Friends