# Message #3353

From: David Vanderschel <DvdS@Austin.RR.com>

Subject: Re: [MC4D] Re: About the number of permutations of MC4D calculating

Date: Sat, 07 May 2016 19:09:38 -0500

I was hoping to spot the error.

Reordering the terms:

B: 96!/(4!)^24 * 96!/(4!)^24 * 48!/(6!)^8 * 96!/(12!)^8 * 64!/(8!)^8 *

(24!*32!)/2 * 2^95 * 2^95 * 2^23 * 3^63 * 64!/2 * 3!^31 * (4!/2)^15

* 16! * 4

S: (96!/24^24)^2 * 48!/(6!)^8 * 96!/(12!)^8 * 64!/(8!)^8 *

(24!*32!/2) * (2^96/2)^2 * 2^24/2 * 3^64/3 * 64!/2 * 6^32/2 * 12^16/3

* (16!/2)

These match up OK except in the last 3 or 4 which can be rewritten:

B: 6^31 * 12^15 * 16!*4

S: 6^32/2 * 12^16/3 * 16!/2

Dividing Smith’s by Balandraud’s:

3 * 4 * 1/8 = 12/8 = 1.5

Unfortunately, that fails to give me a clue about what the mistake might

be. Indeed, it would be hard to explain as a single mistake. I briefly

thought maybe it might be the Klein group on a 4-sticker cubie, but that

would have a larger effect.

Regards,

David V.

On 5/7/2016 3:52 AM, joelkarlsson97@gmail.com [4D_Cubing] wrote:

>

>

> I think it’s time to revive this old thread since I have found that

> there is a difference between the value of the number of

> configurations of the 5^4 calculated by David Smith (in his n^4 paper)

> and the one calculated by Eric Balandraud (at Superliminal). I have

> tested applying Smith’s formula and this gives the same value as Smith

> derives earlier in his paper. They differ by a factor of 1.5 (Smith’s

> value is the greater one) and I think that Balandraud might have

> missed something or simply mistyped the calculation. What do you think?

>

> Balandraud’s calculation (which gives the value at Superliminal):

> 48!/(6!)^8 * 96!/(12!)^8 * 64!/(8!)^8 * (24!*32!)/2 * 3!^31 * 2^23 *

> 64!/2 * 3^63 * 16! *(4!/2)^15 * 4 * 96!/(4!)^24 * 2^95 * 96!/(4!)^24 *

> 2^95

>

> Smith’s calculation (which corresponds to Smith’s formula):

> (16!/2) * (24!*32!/2) * 64!/2 * (96!/24^24)^2 * 64!/(8!)^8 *

> 96!/(12!)^8 * 48!/(6!)^8 * 12^16/3 * 6^32/2 * 3^64/3 * 2^24/2 * (2^96/2)^2

>

> It’s also worth mentioning that Smith’s formula and Balandraud’s

> calculations provide the same value for the 3^4 and 4^4 cube.