Fun with Numbers: The Pan-Prime-Digit Cube Hypothesis

[Joe Horn]

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.

[SlideRule]

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

[Bob Patton]

I think I have found a proof, but the margins here are to small to contain it.

[Dan]

Ah, but then you miss out on the Millennium Prize $1,000,000 award.

[Luigi Vampa]

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

[Gerald H]

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

[Joe Horn]

(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.

[SlideRule]

ALL

for the Latin challenged;

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

BEST!
SlideRule

[rprosperi]

(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"

[Joe Horn]

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."

[Massimo Gnerucci]

(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! :)

[Gerald H]

I draw hope from the "Too soon".

Still working at it.

[AlexFekken]

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.

[AlexFekken]

"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...

[David Hayden]

(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?

[Werner]

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

[David Hayden]

(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!]

[Paul Dale]

Rereading the first message and the base number doesn't have to be pan-digital -- the example of the square is such.

Pauli

[Werner]

(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

[Arno K]

(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