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- What language would they have used

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JohnNichols

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08-02-2020
09:30 AM

547 Views

What language would they have used

I stumbled across this little gem, it is correct - checked it with excel -- did they likely do it in Fortran and what algorithm do you think they used?

This paper sets out some of the findings as well from 1967, this is a big search and overflow would be a real issue

The two are from the aerospace co - why would they be interested in this pure math --

Link Copied

28 Replies

JohnNichols

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08-02-2020
05:30 PM

438 Views

How do I make it threaded to speed up the analysis

```
! Console100.f90
!
! FUNCTIONS:
! Console100 - Entry point of console application.
!
!****************************************************************************
!
! PROGRAM: Console100
!
! PURPOSE: Entry point for the console application.
!
!****************************************************************************
program Console100
implicit none
double precision a(1000), b(1000), count, sum
integer i,j,k,l,g,h
! Variables
! Body of Console100
print *, 'Hello World'
open(1,file="f.out",status="unknown")
count = 1.0
h = 144
do 100 i = 1,h
a(i) = count*count*count*count*count
b(i) = a(i)
write(2,*)i, a(i)
count = count + 1.0
100 end do
do 200 i = 1, h
write(1,*)i
do 300 j = 1, h
do 400 k = 1, h
! write(*,*)i,j,k
do 500 l = 1,h
sum = a(i) + a(j) + a(k) + a(L)
write(1,*)(sum**0.2)
do 600 g = 1, h
if(sum .eq. b(g)) then
write(1,*)">>>>>>>>>>>>>>>>>>>>>>",g, sum
end if
600 end do
500 end do
400 end do
300 end do
200 end do
end program Console100
```

- this i brute force - we should be able to speed it up - optimize operations

mecej4

Black Belt

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08-02-2020
06:47 PM

431 Views

Curiously, the same right hand value, i.e., 144^5, *can* be expressed as the sum of the 5th powers of 5 integers:

a^5 + b^5 + c^5 + d^5 + f^5 = n^5 = p^5 + q^5 + r^5 + s^5

for a = 38, b = 86, c = 92, d = 94, f=134, and n = 144. As stated in the paper, p = 27, q = 84, r = 110, s = 133.

Furthermore, all the numbers a, b, c, d, f and n are even.

A related curious fact: The fifth power of any integer has the same last digit as the number itself does. This property can be used when testing if a given integer is the fifth power of an integer.

Wikipedia has an article on the conjecture, see https://en.wikipedia.org/wiki/Euler%27s_sum_of_powers_conjecture .

According to the article, the truth of the conjecture remains an open question when the exponent k > 5.

P.S. (3 Aug 2020):

Some additional information follows. The paper that JMN attached is one of the shortest published journal articles in existence (it consists of two sentences). A more detailed paper, of which Lander is also an author, is https://www.ams.org/journals/mcom/1967-21-099/S0025-5718-1967-0222008-0/S0025-5718-1967-0222008-0.pd... . In the notation introduced in that paper, the 2-sentence paper announces a (5. 1. 4)<sub>1</sub> "primitive" solution.

The solution that I gave above is a (5. 1. 5) solution, which I obtained from a program in which I started with n = 144 and searched for a few smaller values of n, but it is not primitive, and therefore is not to be found in Table III of the longer AMS paper. If we divide each of the numbers in the equality by 2, the result is the first entry, which IS primitive:

a = 19, b = 43, c = 46, d = 47, f = 67 and n = 72

A distributed computing project targeting sums of sixth powers is described at http://euler.free.fr .

JohnNichols

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08-02-2020
09:19 PM

420 Views

I am of the opinion that the Lander solution actually should include 0 to the 5th paper, it is just the zero on the function can be represented with 4 integers > 0, it is really a zero, was one thought .

So I am not sure Euler was theoretically wrong, just a way of expressing the answer.

It is very slow for even 144.

JohnNichols

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08-02-2020
09:21 PM

417 Views

It is unknown whether the conjecture fails or holds for any value *k* ≥ 6.

We can show it fails, but I doubt we can prove it is true, we cannot represent all the numbers in a computer accurately in a time that can be managed - is a suggestion -- proving it by pure math - not sure.

JohnNichols

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08-02-2020
09:23 PM

416 Views

Actually for k> 6 we should work out an algorithm that finds each zero quickly

JohnNichols

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08-02-2020
09:26 PM

414 Views

And the brute force solution finds 24 solutions for the set 0 -144 run 4 times

in about 10 minutes on a core i5

andrew_4619

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08-03-2020
01:04 AM

405 Views

I would guess looking at number theory will yield some smarter method. But why not to your example with I8 rather that real. Also as a<b<c<d why not have the finding loop i+1, h and j+1, h etc at least then you look at less combos.

Also have found an answer you can exit the test loop rather than continue

JohnNichols

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08-03-2020
09:49 AM

358 Views

1. which is the bigger potential number the i8 or the double precision

2. Yes, lying in bed last night, I was doing the loops in my head, if I follow your idea I will reduce the loop count by the factorial of k-1.

3. I have a spare NUC I can turn the problem loose on or I can use a supercomputer

4. I was then going to turn the problem loose on 6.

JohnNichols

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08-03-2020
09:50 AM

357 Views

I wanted to test it all the way through - so yes I will put a stop in it.

Thanks

JohnNichols

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08-03-2020
01:07 PM

337 Views

Is there anything I can do to optimize the 4 do loops

```
! Console100.f90
!
! FUNCTIONS:
! Console100 - Entry point of console application.
!
!****************************************************************************
!
! PROGRAM: Console100
!
! PURPOSE: Entry point for the console application.
!
!****************************************************************************
program Console100
implicit none
double precision a(1000), b(1000), count, sum
integer i,j,k,l,g,h,m,n,p
! Variables
! Body of Console100
print *, 'Hello World'
open(1,file="f.csv",status="unknown")
count = 1.0
h = 4
m = 1
n = 1
p = 1
k = 0
l = 0
j = 0
sum = 0.0
do 100 i = 1,h
a(i) = count*count*count*count*count
b(i) = a(i)
!write(2,*)i, a(i)
count = count + 1.0
100 end do
i = 0
do 200 i = 1, h
if(i .gt. 1) then
p = i
end if
do 300 j = p,h
if(j .gt. 1) then
m = j
end if
do 400 k = m,h
if(k .gt. 1) then
n = k
end if
do 500 l = n,h
!sum = a(i) + a(j) + a(k) + a(l)
write(1,1000)i,j,k,l, sum
write(*,1001)i,j,k,l
1000 Format(i5,',',i5,',',i5,',',i5,',' f15.1)
1001 Format(i5,' ',i5,' ',i5,' ',i5)
500 end do
400 end do
300 end do
write(*,*)' '
200 end do
end program Console100
```

JohnNichols

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08-03-2020
01:11 PM

335 Views

```
do 200 i = 1, h
! if(i .gt. 1) then
! p = i
! end if
do 300 j = i,h
! if(j .gt. 1) then
! m = j
! end if
do 400 k = j,h
! if(k .gt. 1) then
!! n = k
! end if
do 500 l = k,h
!sum = a(i) + a(j) + a(k) + a(l)
write(1,1000)i,j,k,l, sum
write(*,1001)i,j,k,l
1000 Format(i5,',',i5,',',i5,',',i5,',' f15.1)
1001 Format(i5,' ',i5,' ',i5,' ',i5)
500 end do
400 end do
300 end do
write(*,*)' '
200 end do
```

JohnNichols

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08-03-2020
01:12 PM

334 Views

That is like the old code from the 1960s - you were lucky to understand it at all

JohnNichols

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08-03-2020
01:33 PM

329 Views

It is just a random fluke of nature that the odd one works

The addition of four fifth order functions equated to a single fifth order function is an exercise, see picture in one function almost following the other -- there is an error between them -- I would guess likely Gaussian or log Gaussian and at one point the error = 0

For some reason I could not insert a picture -- the grey is the fifth order function and the two dots are the first two sets of the four additions.

JohnNichols

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08-03-2020
01:46 PM

320 Views

Hp Envy Corei5 65 seconds of cpu time - I wonder how long on the CDC 6600

JohnNichols

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08-03-2020
11:00 PM

303 Views

Found second one - after 4 hours -- 54 168 220 266 - it is the double of the first one,

so next one is likely 108 336 440 532 = 576

next one 216 672 880 1064 = 1152

JohnNichols

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08-03-2020
11:03 PM

300 Views

1. I am of the opinion that lander did not have enough computing time to find the second one-- it took forever,

2. can I represent this number as an integer.

6.34034E+13 |

GVautier

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08-04-2020
07:51 AM

283 Views

JohnNichols

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08-04-2020
08:06 AM

291 Views

I was wrong, there are more, they are not common about 1 in a million combinations to the first 1000, after 12 hours I am 135 - not even cracked the

it appears so far on second glance they are all integer multiples of the first one - or

27 54 81 108 135

27 = 3*3*3

54 = 3*3*3*2

81 = 3*3*3*3

135 = 3*3*3*5

81 252 330 339 - this is the addtion of the first two -

135 420 550 665 - 135 seems weird - it is still going -- at the moment I have it searching to the first 1000 -- I can probbaly do to about 6000 on the rhs before we have an integer fail.

Speed is not a problem - I have spare computers all over the place - some faster than this one --

I want to try 6 next.

I would appreciate the unlimited integer math

There is a pattern to the numbers all based on 3. let us see if we find more

JohnNichols

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08-04-2020
08:09 AM

294 Views

162 504 660 798 just popped up

next one is

189 588 770 931 - should find it in about an hour

JohnNichols

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08-04-2020
08:15 AM

204 Views

144 = 3*3*2*2*2

There is a simple beauty to them ==

Trick is -- are there any others

Guys:

Can you take a look at the code and show me how to thread it or OMP --

Thanks

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