Timeline for Analytic expression for the coefficient of a multivariate polynomial
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Mar 29, 2022 at 22:00 | vote | accept | Fabius Wiesner | ||
| Mar 29, 2022 at 20:33 | answer | added | Peter Taylor | timeline score: 2 | |
| Mar 29, 2022 at 13:34 | comment | added | Brendan McKay | There is also a sum of $O(k^4)$ terms which are products of 7 binomial coefficients $\binom{k}{x}$ for various $x$. That might be fastest of all but I won't do it. | |
| Mar 29, 2022 at 13:04 | comment | added | Brendan McKay | @PeterTaylor Maple is much slower than SageCell for this, but of course custom code is unbeatable. I put together a C program using gmp and it went up to 128 terms in 40 seconds and 200 terms in 6 minutes. The space requirement is k^3 big integers. Indefinitely many terms could be found in little space by Chinese Remainder Theorem. | |
| Mar 29, 2022 at 8:40 | comment | added | Peter Taylor | With Per's approach, 49 or 50 terms in SageCell: online demo | |
| Mar 29, 2022 at 8:20 | comment | added | Peter Taylor | @BrendanMcKay, the obvious Sage two-liner gets 24 terms (25 including $k=0$) in two minutes, but memory usage starts to become a problem. Per's approach would squeeze out a few more terms. Sequence continues 208700059508979095231, 10779043561142979028542, 560841942189830834273092, 29369550998772449215630125, 1546739794102935909806815249, 81869695539756400950043474673, 4352934931813081431729948218755, 232378908415834696591512040624873, 12450814046356118070132295043450105, 669328869650501920320002045039206532, 36090724465969619378056260542868266088, 1951440451620951424228174912475314443293. | |
| Mar 29, 2022 at 3:19 | comment | added | Brendan McKay | 5, 125, 4217, 163373, 6873505, 305304605, 14090295602, 669067354925, 32476460956025, 1604222480193625, 80380036959149680, 4075387263236663069 Special code could quickly find many more terms, perhaps more than 100. | |
| Mar 27, 2022 at 8:32 | comment | added | Brendan McKay | 5, 125, 4217, 163373, 6873505, 305304605, 14090295602, 669067354925, 32476460956025, 1604222480193625. Probably the asymptotic behaviour can be found. | |
| Mar 27, 2022 at 0:13 | comment | added | Peter Taylor | The OEIS Superseeker has no suggestions whatsoever, which is fairly discouraging. | |
| Mar 26, 2022 at 18:56 | history | edited | Fabius Wiesner | CC BY-SA 4.0 |
Added simpler case.
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| Mar 26, 2022 at 18:11 | comment | added | Per Alexandersson | Note that if you can solve it without the last factor, then you can solve it with the last factor as well. The former starts as {4, 84, 2353, 74644, 2570504, 93417141, 3526418676, 136938227092} if my Mathematica is correct. | |
| Mar 26, 2022 at 17:20 | comment | added | Dattier | $z_1=x,z_2=x^{12k},z_3=x^{(12k)^2}$ you must find the coeff of $x^{k+k\times 12k+k\times (12k)^2}$ | |
| Mar 26, 2022 at 16:51 | comment | added | Ben McKay | Taylor coefficient doesn't help? | |
| Mar 26, 2022 at 16:44 | history | asked | Fabius Wiesner | CC BY-SA 4.0 |