The "beast" in full beauty:

I used the 4-cable (knot) with (blackboard) framing 16 and pattern the
braid [−2−3−1−2−1], i.e., the unframed cable
with 4-braid pattern [(123)^{64}−2−3−1−2−1]
(=[1(123)^{62}] up to conjugacy; "±*i*" stands for σ_{i}^{±1}).

The polynomial until *m*-degree 4 is given below,
in the format where the first row records the *m*-degrees
(0 to 4) and the first two integers in every next row the
*l*-degrees for given (even) *m*-degree. (Clearly, powers
of both *l* and *m* must be even in every monomial.)

261 1057125 0 4 220 252 107 -22019 338106 902878 -718680 -3767346 -2410797 2607686 3813661 885472 -889806 -543819 -41263 37391 9639 739 16 218 254 112 -224220 13177877 -77912165 -158848554 435850104 1093610477 174760000 -1319003747 -1179669367 -56397912 376191658 174320733 8332538 -12598957 -3396172 -337832 -12978 -94 218 256 -347866 99428343 1674801372 -1409217546 1414376011 -2040756822 -523606069 -1649687580 -734358094 -1114907086 -2123161090 997162016 1942589839 -476514278 1750181528 452980766 47861085 2234154 35523 76

Here, like elsewhere on my pages, the "mirrored" (with
*l* and *l*^{−1} exchanged)
Lickorish-Millett convention is used.
The coefficients were calculated as 4-byte machine integers (integers
modulo 2^{32} taken between −2^{31}
and 2^{31}−1).

To rule out braid index 5, by MFW, we need a term in *l*-degree
(at least) 256. The problem with this knot is that such a term occurs
only for *m*-degree 4. For all its "siblings", degree at most 2
sufficed. At a diagram of 261 crossings, the difference between
*m*-degrees 2 and 4 is one between 6 weeks and 13 years of CPU time.

You can see the hardware I used on this photo of my desk (Oct 2022):

I had a *computation server* (right, front) with 20 3.4-GHz
CPUs and another *desktop* (left, back) with 8 4.2-GHz CPUs.
This is a solid equipment of the state of the art of early 2010s,
so nothing impressive. I requested no supercomputer timeslots or the
like. I had more hardware available, but I very deliberately tried to
limit myself to these 2 machines. (I did use my – also 10-year
old – laptop for consistency test calculations, though.)

With the parallelized truncated cabled MFW, the calculation took quite exactly half a year, between 6 Nov 2022 and 5 May 2023. (My estimates show that on the computation server without parallelizing, i.e., on a single CPU, I would have needed about 13.4 years.)

The truncation algorithm yields lower *m*-degree terms essentially
for free, which can be calculated independently; this gives
a consistency test.

Another "checksum" test comes from
the cabling property of the Alexander-Conway polynomial ∇.
It can be calculated pretty (at least far more)
easily that the *z*-degree 4 coefficient of ∇ is
2,284,396. This must (and does) appear (modulo 2^{32})
as the alternating sum of the coefficients in the last row.
The first coefficient in the row (third number)
must be taken with '−' sign, as the sum of *l*-degree
(218) and *m*-degree (4) of its monomial is not divisible
by 4. (This checksum test works for lower *m*-degrees as well,
with degree 0 yielding alternating sum 1 and degree 2 yielding 2705.)

I cannot be certain that 3-cable HOMFLY fails, but the terms I'd need to compute promised less for their cost than the 4-cable. (2-cable is way too helpless.)

Also, I added 5 (negative) crossings to get a 4-cable knot basically because this was technically my laziest way to produce a cable knot diagram. (I used it for 4-cabling some of the other knots as well.) Of course, I could have added only 3 pattern crossings, or none at all and treated a cable link. But with fewer crossings, these diagrams have a larger writhe, which ostensibly impacts a skein method (I use). So it is not very clear which would have been faster. And trying to experiment with multiple cable patterns is hardly reasonable at this scale of complexity.

I hope this gives a bit of insight into how I work, and in particular draws away from the impression that I am a computer maniac. (This conclusion is corroborated by a very respectable person who even recently said that I live in the "stony age", as I went to the railway station to buy tickets and do not use an app.)

I would be interested if alternative means (Khovanov-Rozansky?) can simplify this great ordeal.

My experience with determining the braid index somewhat suggests the following very "wild" guess, which I use the occasion to forward here:

**Conjecture.**

(a) The *n*-cabled MFW for all *n* will yield the
braid index for every knot.

(b) For every *n*, there is a knot whose braid index is not
determined by *n'*-cabled MFW for *n'*≤*n*.

Alexander Stoimenow (May 17, 2023),