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Old 2022-05-12, 23:43   #12
marcokrt
 
May 2022

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Since we have two arguments leading to opposite conclusion, we could argue which is the strongest one of the two.
Now, if I should bet (and I do not ever bet), I would say that charibdis' one should be more solid than my empirical evidence based on the range (2, 10^34] or (2, 10^36]... but let us consider a really random similar sequence (I swear, this is the first one which comes to my mind and the most similar that I know by far) in order to test if a perfect cube can occur in this range: perfect powers in A030299, since the sequence A352329 from the OEIS shares the first 36 terms with A353025 and the first one which is not equal is A352329(37) = 1234608769 <> 14102987536 = A353025(37).
Well, we can easily find that there is a perfect cube in A030299 which is pretty close to A352329(37): 8096384512=2008^3, belonging to A030299.

Thus, we have that A353025 and A352329 are the same sequence up to 923187456 and then the generating sequence A030299 of A353329 has a perfect cube which is smaller than 1.5*10^10, while the generating sequence A352991 of A353025 hasn't any cube up to 10^36.

Just a fact, which doesn't show that the empirical argument is better than the others, but IMHO it says something at the end.

Last fiddled with by marcokrt on 2022-05-12 at 23:47
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Old 2022-05-12, 23:46   #13
charybdis
 
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Quote:
Originally Posted by marcokrt View Post
My point is that we do not know whether or not there are additional constraints (unknown by us right now) which could drop the distribution of cubes or prevent a perfect power above two to be a term of A353025, but (since we don't know any of these hypotetical unknown constraints)
Agreed. My point is that these unknown constraints are extremely unlikely to exist. Off the top of my head, I can't think of a single example of a problem like this where the "correct" heuristic suggests infinitely many X ought to exist, but it turns out no X exists and the proof isn't extremely simple. (Even Fermat's Last Theorem follows this rule! n=2 we naively expect infinitely many solutions, n>=3 we naively expect finitely many...)

Quote:
Moreover, there are not sum(j=1, n)j! terms up to the biggest permutation of 1_2_3_..._(n-1)_n (but less than that), since the same number can occur from different permutations (just to point out it, but this is a minor argument).
Yes, I noticed this too. I didn't point it out because it should be clear that the proportion of permutations that are "duplicates" like this is vanishingly small.
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Old 2022-05-12, 23:55   #14
charybdis
 
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Quote:
Originally Posted by marcokrt View Post
but let us consider a really random similar sequence (I swear, this is the first one which comes to my mind and the most similar that I know by far) in order to test if a perfect cube can occur in this range: perfect powers in A030299, since the sequence A352329 from the OEIS shares the first 36 terms with A353025 and the first one which is not equal is A352329(37) = 1234608769 <> 14102987536 = A353025(37).
Well, we can easily find that there is a perfect cube in A030299 which is pretty close to A352329(37): 8096384512=2008^3, belonging to A030299.

Thus, we have that A353025 and A352329 are the same sequence up to 923187456 and then the generating sequence A030299 of A353329 has a perfect cube which is smaller than 1.5*10^10, while the generating sequence A352991 of A353025 hasn't any cube up to 10^36.
A030299 will produce more cubes because the "carrying" ensures that the numbers are smaller and thus more likely to be cubes. Adding two digits on instead of one each time we add a number to the permutation has a big effect - we have to wait longer for the superexponential growth of n! to take effect.
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Old 2022-05-13, 00:28   #15
marcokrt
 
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Quote:
Originally Posted by charybdis View Post
A030299 will produce more cubes because the "carrying" ensures that the numbers are smaller and thus more likely to be cubes. Adding two digits on instead of one each time we add a number to the permutation has a big effect - we have to wait longer for the superexponential growth of n! to take effect.

Thank you for this very enjoyable discussion!
Absolutely agree about the most reasonable guess on the existence of the cubes... here we are almost on the reverse side of Tao's "almost-proof" of Collatz Conjecture, since mine is "probably" false; so, the most interesting part would be to wait for a formal disproof... around the 24th century maybe
As a memory of the sequences A352991 and A353025, I'll still call it "Conjecture" untill somebody will formally disprove it, even if now we know that it wouldn't probably hold up to very large numbers as 10^80 or so, since it needs that some kind of unknown (heavy and general) constraint occurs along the way.
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Old 2022-05-14, 13:28   #16
Dr Sardonicus
 
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If I understand the problem correctly, you could (in theory) form your "permuted" integers as follows:
  1. Form a vector of length n whose i-th entry is i.
  2. Convert the entries to strings (of decimal digits).
  3. Permute the entries
  4. Concatenate the permuted entries
  5. Convert the resulting string to an integer in base ten.

If the resulting integer is to be a cube, it has to be congruent to 0, 1 or 8 (mod 9).

However, any concatenation of the integers 1 to n is congruent to the sum of the integers 1 to n, or n(n+1)/2 (mod 9).

This gives the requirement that n be congruent to 0, 1, 4, 7, or 8 (mod 9). (No number n(n+1)/2 is congruent to 8 (mod 9).)

This is of course only a "constant factor" thing. I find the argument persuasive that, when n is "large enough" (and the possibility of cubes is not ruled out by congruences etc) then there will very likely be cubes among the "concatenated permuted" integers 1 to n.

I find the argument similarly persuasive for higher powers.

I'd call the question an "open question" rather than a "Conjecture." It's a numerical curiosity, but AFAIK does not have any theoretical significance.
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Old 2022-05-14, 16:52   #17
marcokrt
 
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Quote:
Originally Posted by Dr Sardonicus View Post
If I understand the problem correctly, you could (in theory) form your "permuted" integers as follows:
  1. Form a vector of length n whose i-th entry is i.
  2. Convert the entries to strings (of decimal digits).
  3. Permute the entries
  4. Concatenate the permuted entries
  5. Convert the resulting string to an integer in base ten.

If the resulting integer is to be a cube, it has to be congruent to 0, 1 or 8 (mod 9).

However, any concatenation of the integers 1 to n is congruent to the sum of the integers 1 to n, or n(n+1)/2 (mod 9).

This gives the requirement that n be congruent to 0, 1, 4, 7, or 8 (mod 9). (No number n(n+1)/2 is congruent to 8 (mod 9).)

This is of course only a "constant factor" thing. I find the argument persuasive that, when n is "large enough" (and the possibility of cubes is not ruled out by congruences etc) then there will very likely be cubes among the "concatenated permuted" integers 1 to n.

I find the argument similarly persuasive for higher powers.

I'd call the question an "open question" rather than a "Conjecture." It's a numerical curiosity, but AFAIK does not have any theoretical significance.
I agree, the "Conjecture 1" name comes from the preprint and the OEIS sequence A353025, but I'll clarify this in the new version which I'm going to update in a couple of weeks (at most).
BTW, we still have a more interesting question to answer:
"Is Kashihara's Conjecture true according to the same probabilistic argument?".
Here is the probabilistical upper bound which I have found for the existence of a perfect power belonging to A001292 (Kashihara's Conjecture says that no term of A001292 which is above 1 can be a perferct power); knowing that the smallest counterexample have necessarily to be above 10^1235 we have:

2*(sum(k=2, infinity)(sum(j=308, infinity)((((10^((j*4+3)/k)) - 10^((j*4+2)/k))/(9*10^(j*4+2)))*(j*4+3)))),

since 308*4+3=1235 (we need to add at least 4 digits to go from A001292(n) to A001292(n+j*4+3)), while I am not sure about the *2 factor which is just an upper bound that I've added in order to take into account the increased chance that the set of the circular permutations of 1_2_3_4_..._447_448 contains any perfect power (i.e., digital root(123...447448)=1, so that its chance should be 9/5 times higher than the expected perfect square ratio in [10^1235, 10^1236-1]).

Now, if the given probabilistic bound holds/is correct, we have that


2*(sum(k=2, infinity)(sum(j=308, infinity)((((10^((j*4+3)/k)) - 10^((j*4+2)/k))/(9*10^(j*4+2)))*(j*4+3)))) is roughly (2/9)*(sum(j=308, infinity)((((10^((j*4+3)/2)) - 10^((j*4+2)/2)/(10^(j*4+2)))*(j*4+3)))) < 10^(-1000000000)

and this would make Kashihara's conjecture 99.9999...% true, even if it is still waiting for a formal proof.

My concern is about the sequences A001292 and A352991:
Let A001292 or A352991 be given and assume that we select the string 123...m such that 123...m==1(mod 9). Is there a more accurate way to estimate the provided multiplicative factor for any given integer value of the exponent (i.e., charybdis wrote that it would be 6 if we take into account the permutations of 1_2_3_..._42_43 following the A232991 permutation rule, but I cannot understand how this value can be determined, since there are 3 out of 9 congruence classes (modulo 9) such that the digital root of 123...m is 1)?


Thanks in advance to everybody!

Last fiddled with by marcokrt on 2022-05-14 at 17:38
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Old 2022-05-19, 16:46   #18
marcokrt
 
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Just to say that the link in the opening post of the present thread is now unavailable, since the current version of the preprint can be read here: https://doi.org/10.13140/RG.2.2.30313.77921/3


Basically, the last remark synthesize the probabilstic argument and extends it to Kashihara's conjecture too... so, after having checked every term up to 10^1235, we are persuaded that it would be almost 100% true under the discussed assumptions, but we still cannot prove it.
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