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Old 2022-05-14, 16:52   #17
marcokrt
 
May 2022

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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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