Timeline for Number of matrices with unit determinant and fixed sum of elements
Current License: CC BY-SA 4.0
22 events
| when toggle format | what | by | license | comment | |
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| Mar 2, 2023 at 8:05 | comment | added | Brendan McKay | @PavelGubkin Agreed. | |
| Mar 2, 2023 at 6:17 | comment | added | Pavel Gubkin | I get 0, 0, 0, 12, 144, 864, 3552, 11592, 32832, $\ldots$ for $a^{(4)}_n$ and 0,$\ldots$, 0, 48, 288, 1440, 4320, 12816, 30312,$\ldots$ for $b^{(4)}_n$ (the first non-zero is on the 19-th position). | |
| Mar 2, 2023 at 4:19 | comment | added | Brendan McKay | Please find a few numbers for $4\times 4$ matrices, so as to check mine. An effort is underway to get these numbers into OEIS. | |
| Mar 2, 2023 at 1:16 | answer | added | Brendan McKay | timeline score: 4 | |
| Mar 2, 2023 at 0:12 | comment | added | Brendan McKay | I agree with your new numbers. Also, $a_{101}=48283272$ and $b_{110}=34841844$. | |
| Mar 1, 2023 at 19:57 | comment | added | Pavel Gubkin | @BrendanMcKay, you are right! I had a problem with the accuracy of determinant function from numpy. | |
| Mar 1, 2023 at 19:52 | history | edited | Pavel Gubkin | CC BY-SA 4.0 |
The first numbers were previously computed incorrectly.
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| Mar 1, 2023 at 17:57 | comment | added | Pavel Gubkin | @BrendanMcKay, oh, will check my computations... | |
| Mar 1, 2023 at 17:25 | answer | added | Andy | timeline score: 3 | |
| Mar 1, 2023 at 13:17 | comment | added | Brendan McKay | Here is a proof that $b_{13}=49$ is wrong. Consider such a matrix $A$. Since $|A|=1$, the columns are distinct. Consider the set of 6 matrices which have the same columns as $A$ in some order. They are all distinct and half have determinant $+1$ and half $-1$. So $b_{13}$ must be a multiple of 3. Similarly for $a_8$. | |
| Mar 1, 2023 at 13:03 | comment | added | Brendan McKay | I don't the same numbers as you. For $a_n$ starting at $n=3$, I get 3, 18, 54, 126, 261, 432, 783 (differing starting at $n=8$. For $b_n$ starting at $n=11$, I get 9, 18, 72, 108, 234, 360 (differing starting at $n=13$). Of course it might be my bug. | |
| Mar 1, 2023 at 7:16 | comment | added | Pavel Gubkin | @BrendanMcKay, thanks, my bad! Now it should be correct | |
| Mar 1, 2023 at 7:15 | comment | added | Pavel Gubkin | @GerryMyerson, $n$ is a sum of entries, you are correct, it is now noted in the question. For $2\times 2$ matrices the answer is not increasing, so we can also expect it here. As you can see for other primes $p$ ($13, 17, 19$) $b_p$ is significantly larger than $b_{p - 1}$ and almost $b_{p + 1}$. The answer could be of the form $O(\phi(n)\cdot P(n))$ with some polynomial $P$ | |
| Mar 1, 2023 at 7:05 | history | edited | Pavel Gubkin | CC BY-SA 4.0 |
added 41 characters in body
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| Mar 1, 2023 at 5:15 | history | edited | Pavel Gubkin | CC BY-SA 4.0 |
edited body
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| Mar 1, 2023 at 0:56 | comment | added | Brendan McKay | And the number of upper triangular matrices with 1s on the diagonal and a composition of $n-3$ above the diagonal is not ${}\ge Cn^3$. It is $\Theta(n^2)$. Once you choose two of those entries, the third is forced. | |
| Mar 1, 2023 at 0:11 | comment | added | Brendan McKay | The number of compositions of $n$ with 9 parts is $O(n^8)$. | |
| Feb 28, 2023 at 22:24 | comment | added | Gerry Myerson | $n$ is much used but never defined. I take it $n$ stands for the sum of the entries? Should I be surprised that $b_n$ is not an increasing sequence? | |
| Feb 28, 2023 at 18:37 | comment | added | Pavel Gubkin | @RobertIsrael, yes, thank you. Added it now | |
| Feb 28, 2023 at 18:36 | history | edited | Pavel Gubkin | CC BY-SA 4.0 |
added 8 characters in body
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| Feb 28, 2023 at 18:35 | comment | added | Robert Israel | You want the entries to be integers? | |
| Feb 28, 2023 at 18:32 | history | asked | Pavel Gubkin | CC BY-SA 4.0 |