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Let $p$ be the vector of dimension $n^2$ consisting of ordered prime numbers i.e. $p= [ 1 \ 2 \ 3 \ 5 \ 7 \ldots]^T$ and $A$ be the matrix of dimension $n\times{n}$ constructed with this vector by the following way:

  • the first column of $A$ is the first $n$ prime numbers from vector $v$

  • the second column of $A$ is the next $n$ prime numbers from vector $v$

  • etc.

Examples of such matrices:
$\begin{bmatrix} 1 & 3 \\ 2 & 5 \\ \end{bmatrix}$, $\begin{bmatrix} 1 & 5 & 13 \\ 2 & 7 & 17 \\ 3 & 11 & 19 \end{bmatrix}$, $\begin{bmatrix} 1 & 7 & 19 & 37 \\ 2 & 11 & 23 & 41 \\ 3 & 13 & 29 & 43 \\ 5 & 17 & 31 & 47 \end{bmatrix}$, $\dots$

Question:
is it true that for any $n$ $ \ $ rank$(A)=n$ (i.e. columns of $A$ are linearly independent) or for some $n$ the statement above is not true?

Widawensen
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    Usually, the primes start with the number $2$ – Peter Dec 07 '16 at 11:04
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    Probably, the matrices all have rank $n$, but a proof will be very difficult – Peter Dec 07 '16 at 11:06
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    No matter, whether we start with $1$ or $2$, for $1\le n\le 200$ , we get a matrix with rank $n$ – Peter Dec 07 '16 at 11:15
  • @Peter Thank you Peter for confirmation that the conjecture is true for $n<201$, ...I was also thinking whether to start this vector $p$ from 1 or 2 .. but to the $n=4$ it was satisfied for both cases... and you have confirmed now that it is true up to the $n=200$. – Widawensen Dec 07 '16 at 17:43
  • If someone kind can confirm conjecture for some greater n>200 please give information here to the other users, I have no tools to operate with such big matrices built from so many primes.. – Widawensen Dec 07 '16 at 19:34
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    @Peter, Since you have a code to check this, were you able to check that the determinants are positive and increasing? It would be nice to look for some pattern if there exists any – Sungjin Kim Dec 08 '16 at 04:15
  • @i707107 Yes, it would be very interesting... determinants for small dimensions matrices, I've checked, are also non-zero if we start the vector $p$ not necessary from the first number, but from some following e.g 3, 7, 11 etc.. but in this case determinants are changing a sign.. – Widawensen Dec 08 '16 at 07:00
  • The late result from Robert Israel https://oeis.org/A067276/b067276.txt. ( matrix starts at 2) The number of digits for the determinant seems to increase almost in linear way. – Widawensen Jul 13 '17 at 18:15
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    The log of the absolute values of the determinants $\ln |d(n)|$ seem to agree roughly to $\frac \pi4 \cdot (\ln |d(n)|)^{0.81} \approx n$ using dimension $n$ up to 100 so the growth of the log would rather appear nonlinear. – Gottfried Helms Sep 13 '18 at 22:57
  • @GottfriedHelms Very interesting result, how it was found? – Widawensen Sep 14 '18 at 08:14
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    I used Pari/GP and did the matrix-calculations. Then tried some transformations on the values of the determinants and the $n$-(matrixsize) parameter and used Excel to locate some trend. That given estimate was based on values for $n=2$ to $n=100$. Later I found a list of values up to $n=459$ provided by Robert Israel and accessible via OEIS. That indicates a slight adaption to a better formula - didn't look at it as intense as before, but again suggested non-linear connection. Perhaps there is something involved with binomials in the exponent... – Gottfried Helms Sep 14 '18 at 08:38
  • @GottfriedHelms Interesting that 0.81 is "almost" equal to $\pi/4$ .. – Widawensen Sep 14 '18 at 11:07
  • @ widawensen: well, sure. But the additional data by R. Israel should be taken into calculation and a better guess for the function should be done. I stepped out after I'd seen, that it is surely non-linear (because linearity was suggested by one earlier comment); after that I tried a handful of types of trends getting that one which pleased me on visual impression. It surely needs some more serious approach with more shapes of the trend-curve... – Gottfried Helms Sep 14 '18 at 11:15
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    Here is an idea I searched on further: https://qr.ae/TJr1AY &
    https://oeis.org/A067276     – Matcha Latte Jan 10 '20 at 07:46
  • @Praskovya2.718281828 Thank you for interesting links :) – Widawensen Jan 10 '20 at 12:38
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    I would perhaps try an alternative method of filling out to higher matrices. For example there are many ways to fill out that lead to the previous matrix being a submatrix in the upper left corner of the next matrix. That sets you up to be able to use induction potentially, but the proof is still not pretty. – EffingLoveMath Jul 02 '21 at 23:28
  • 1 isn't a prime number because it only has 1 factor. A prime number must have two factors - 1 and itself. – MathGeek Aug 25 '21 at 13:36

1 Answers1

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As far as I can tell, this conjecture holds but there doesn't appear to be any patterns that help us with computation.

By bruteforcing the determinant, we find that this holds up to n = 459 (based on the following OEIS documentation: https://oeis.org/A067276/b067276.txt). However, the magnitude of determinants between n = i, i+1 are not necessarily increasing for every such i (e.g. determinant of n = 27 > determinant of n = 28).

Bryce
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