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A "pan-prime-digit number" is a natural number containing all four prime digits (2, 3, 5, and 7) in any order, one or more times each, but no other digits. The smallest pan-prime-digit number is 2357, which also happens to be a prime number, but there are obviously infinitely many pan-prime-digit numbers, and probably infinitely many prime ones. I'm pretty sure that the smallest pan-prime-digit number which is a perfect SQUARE is 23377225 (equal to 4835^2). There are probably infinitely many pan-prime-digit squares.

However, it is my hypothesis that there is ONLY ONE pan-prime-digit CUBE. I would be delighted beyond words if anybody could either prove (mathematically) or disprove (by counterexample) this hypothesis. Needless to say, finding the one known pan-prime-digit cube is left as a mini programming challenge. This posting appears in the "Not remotely" forum because no current HP programmable calculator is fast enough to find the number in a reasonable amount of time; it's surprisingly large.
Perhaps the following url's ...
a) Prime Curios!- 7
b) Prime Curios!- 89
c) An ancient Greek algorithm could reveal all-new prime numbers
... inspire / edify the pure math curious / challenged (myself included). Interesting read, Joe, thanks!

BEST!
SlideRule
I think I have found a proof, but the margins here are to small to contain it.
Ah, but then you miss out on the Millennium Prize \$1,000,000 award.
After three days trying to prove the 'Horn Conjecture', I throw the towel ;0) I think I will devote myself to implement symplectic integration in Free42, in order to start searching for Planet Nine. The latter is going to be much easier for good >D
To avoid suspense: Horn's Conjecture is WRONG.

After prolonged hesitation & cogitation I decided to join the throng of adepts attempting to solve the question of Joe's conjecture.

First some general observations:

1 The ancient Greeks have very little to say: they were not interested in digits in a positional representation of numbers, they were interested in the properties of numbers;

2 The proof below exceeds the limits of the margin, nevertheless I will publish the complete proof to expose it to the court of my peers, superiors & inferiors;

3 I am not interested in financial gain.

An approach is to convert a problem in multiplicative number theory (structure of x^3) to one of additive number theory (sum of consecutive odd integers).

For integer input N the programme below returns N & a list of consecutive odd integers that sum to n^3:
Code:
``` ::   CK1&Dispatch   BINT1   ::     %ABSCOERCE     DUP#0=csedrp     Z0_     DUP1LAMBIND     #1-     DUPDUP     #*     #+     #1+     DUP     1GETLAM     DUP     #+     #+SWAP     DO     INDEX@     FPTR2 ^#>Z     BINT2     +LOOP     1GETLAM     {}N     1GETABND     FPTR2 ^#>Z     SWAP   ; ;```
However, having considered the bijection of cubes & sums of consecutive odd integers I concluded this was a dead-end.

& similarly hopeless is a proof in the traditional sense, as base 10 representation of a number tells us more about 10 than of the number represented. Number theory speaks of the properties of numbers, eg

3153023022 base 7 is an odd number,

& this remains true if converted to base 10 or any base.

So I arrived at a heuristic proof.

What is the probability of there being exactly one cube, call it H, with decimal digits exclusively 2,3,5,7 & each of these digits appearing in the representation?

Difficult to say, but surely very small.

The probability can be further diminished by adding that Joe found H.

That Joe found H amongst the infinitude of cubes can only indicate that there must be a large number of cubes with the required property.

Indeed, using the maximum likelihood hypothesis, the greatest probability of Joe finding H occurs when there is an infinity of such cubes.

Proving, to all intents & purposes, that there is an infinity of the proposed cubes.

QED
(08-06-2017 09:47 AM)Gerald H Wrote: [ -> ]To avoid suspense: Horn's Conjecture is WRONG. ... I arrived at a heuristic proof.

What is the probability of there being exactly one cube, call it H, with decimal digits exclusively 2,3,5,7 & each of these digits appearing in the representation?

Difficult to say, but surely very small. [That's not zero. -jkh-]

The probability can be further diminished [to zero? -jkh-] by adding that Joe found H.

That Joe found H amongst the infinitude of cubes can only indicate that there must be [WRONG -jkh-] a large number of cubes with the required property. [No, it indicates that there MIGHT be MORE THAN ONE. That's all. -jkh-]

Indeed, using the maximum likelihood hypothesis, the greatest probability of Joe finding H occurs when there is an infinity of such cubes. [That makes it more likely, but it is not necessary. In other words, if my hypothesis were correct, would that PREVENT me from finding H? Of course not. -jkh-]

Proving [WRONG. -jkh-], to all intents & purposes [AHA! There it is. -jkh-], that there is an infinity of the proposed cubes.

QED [ <-- Too soon. -jkh-]

See my comments inserted above, inside [ square ] brackets.

As mentioned in the original posting, a counterexample would suffice to disprove the conjecture. The above rationale explains why the conjecture is probably false, but it does not disprove it.

FWIW, if the above were a valid proof, then it would also prove the existence of infinitely many perfect square repunits, since 1 itself is certainly one such. But there are no others. Therefore the "proof" is invalid. QED.
ALL

for the Latin challenged;

QED = quod erat demonstandum = thus it has been demonstrated.

BEST!
SlideRule
(08-06-2017 10:06 PM)SlideRule Wrote: [ -> ]QED = quod erat demonstandum = thus it has been demonstrated.

Funny, I was taught it means "Quit, Everything is Done" or "Quit, Explanation is Done"
Singer-songwriter Thomas Dolby's 1988 song "Airhead" includes the lyric, "Quod erat demonstrandum, baby," referring to the self-evident vacuousness of the eponymous subject; and in response, a female voice squeals, delightedly, "Oooh... you speak French!"

— excerpt from the Wikipedia article about "Q.E.D."
(08-07-2017 03:38 AM)Joe Horn Wrote: [ -> ]Singer-songwriter Thomas Dolby's 1988 song "Airhead" includes the lyric, "Quod erat demonstrandum, baby," referring to the self-evident vacuousness of the eponymous subject; and in response, a female voice squeals, delightedly, "Oooh... you speak French!"

— excerpt from the Wikipedia article about "Q.E.D."

Oh, my poor lingua mater! :)
I draw hope from the "Too soon".

Still working at it.
Following the same logic:

By the prime number theorem, the one that gives the asymptotic distribution of prime numbers, the probability that any positive integer less than N is a prime number goes to 0 (as N goes to infinity). Hence, the probability of any positive integer being a prime number is zero. Therefore there are no prime numbers other than the ones that have already been discovered. QED.
"no current HP programmable calculator is fast enough to find the number in a reasonable amount of time"

Are you sure? Perhaps you are right, but there are some obvious ways to cut down the number of required calculations quite a bit, as compared to a naive brute force attack. E.g. you could calculate the number of different values of n^3 mod 1000 (and then mod 10^6, then mod 10^9) that only contain the required digits. I would expect that that could narrow down the search quite a bit...
(08-07-2017 10:50 AM)AlexFekken Wrote: [ -> ]there are some obvious ways to cut down the number of required calculations quite a bit, as compared to a naive brute force attack. E.g. you could calculate the number of different values of n^3 mod 1000 (and then mod 10^6, then mod 10^9) that only contain the required digits. I would expect that that could narrow down the search quite a bit...

[ Edit: I mistakenly thought that both the number N and the cube N^3 had to be pan-prime-digital. In reality only N^3 must be. The description below reflects my misunderstanding. The basic approach can still be applied to the problem though.]

That's the approach I took last night with a C++ program and BigInt library. Multiplying two numbers with N least significant digits will result in the N least significant digits in the product, regardless of what the more significant digits are. So if you start with the pan-prime-digits and cube them. You find that for any number N ending in 2, N^3 will end in 8, so it can't be a pan-prime-digital cube.

Thus any pan-prime-digit cube must end in 3, 5, or 7.

Next you prepend 2, 3, 5, and 7 to these suffixes and test them. Here you find that any pan-prime-digit cube must end in 25, 33, 25, 37, 53, 55, 75 or 77. The program repeats the process until there are no more suffixes. I hoped that the number would eventually shrink. Boy was I wrong.

My program ran all night. When I stopped it, it was checking 10 million+ 35-digit suffixes and hadn't found a solution yet. I was running it on a slowish computer at work. I'll try running it on my laptop which has more horsepower.

Joe can you confirm that the answer you found is more than 35 digits?
Free42 (what else?) is fast enough. Well, still a couple of hours.
1 405 349 897 ^3 = 2 775 577 757 352 755 375 573 357 273

Cheers, Werner
(08-07-2017 07:57 PM)Werner Wrote: [ -> ]1 405 349 897 ^3 = 2 775 577 757 352 755 375 573 357 273
I don't think this counts. The input number must be pan-digital also. This one contains 0, 1, 4, 8, and 9, and it lacks 2.
[Edit: I see that I'm wrong. The input number does NOT need to be pan-prime-digital. Back to the drawing board!]
Rereading the first message and the base number doesn't have to be pan-digital -- the example of the square is such.

Pauli
(08-07-2017 07:09 PM)David Hayden Wrote: [ -> ]Thus any pan-prime-digit cube must end in 3, 5, or 7.
Or 8.
So that's only 4/10 numbers to test.
If you take the two last digits, it's 24/100

Cheers, Werner
(08-08-2017 06:38 AM)Werner Wrote: [ -> ]
(08-07-2017 07:09 PM)David Hayden Wrote: [ -> ]Thus any pan-prime-digit cube must end in 3, 5, or 7.
Or 8.
So that's only 4/10 numbers to test.
If you take the two last digits, it's 24/100

Cheers, Werner

This is what I noticed, too. So I wrote a program for the Prime using lists of usable endings , ran out of memory after the last 6 digits were checked. Now I try to find an improvement.
Arno
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