[seqfan] Re: error in sequence?

[Christian G.Bower]

Dear seqfans,

I composed this reply to Jud's post last night. Since then there
have been replies covering much of what I've written, but I think there
is enough here that hasn't been covered to make this worth posting.

Jud McCranie wrote:
> I'm wondering if there is an error in A6550.
>
> %S A006550 1,10,57,234,770,2136,5180,11292,22599,42190,74371,124950,201552,
> %T A006550 313964,474510,698456,1004445,1414962,1956829,2661730,3566766
> %N A006550 n-coloring a cube.
> %D A006550 L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 254.
>
> This corresponds to (n^6 - 3n^4 + 48n^3 - 58n^2 + 36n)/24.
>
> I don't have the reference available, but I interpret it as the number of
> distinct ways to color a cube with n colors.  Rotations that give the same
> coloring are counted as equivalent.
>
> I get 1, 10, 57, 240, 800, 2226, 5390, 11712, ...
>  (n^6 + 3n^4 + 12n^3 + 8n^2)/24
>
> Can someone tell me if there is an error in the source, am I interpreting
> the problem incorrectly, or what?  Thanks!

Here is some background to help interpret the problem.

Suppose b_n is a sequence describe the number of ways to color some
object in n colors where all n colors must be used.

Then a_n=C(n,0)*b_0 + C(n,1)*b_1 + ... + C(n,n)*b_n is the number of ways
to color the object with n colors where not all the colors need to be
used. (C(n,k) is the binomial coefficient n!/(k!*(n-k)!).)

The operation to go from b to a is known as the binomial transform in
EIS-land.

Anyway, if A006550 is the "a" sequence, the "b" sequence is:
1 8 30 62 75 30 0 0 0...

If Jud's sequence is the "a" sequence, the "b" sequence is:
1 8 30 68 75 30 0 0 0...

So there is a disagreement about how many ways to 4-color a cube where
all 4 colors must be used.

Well, I've been trying to count the ways myself and I can't seem to
get the same answer twice in a row, but based on my latest count, I
would agree with Jud's claim that A006550 is in error.

I'll diagram as if I folded the cube out into a "T"

0
0 0 1 2
3

I count 8 colorings to this pattern
(4 choices of the 3-fold color and 2 orientations for the remaining
3 colors)

1
0 0 0 2
3

12 of these.
(4 choices of the 3-fold color, 3 choices for which pair of the
remaining colors are opposite one another.)

2
0 0 1 1
3

6 of these

2
0 1 0 1
3

6 of these

1
0 1 0 2
3

12 of these

1
0 0 2 1
3

12 of these

1
0 0 1 2
3

12 of these

for a total of 68.

Maybe someone should check this more carefully though. I sure
don't trust myself.

Here's another sequence that claims to count the ways to
color a cube:

%I A006529 M4717
%S A006529 0,1,10,57,272,885,2226,4725,8912,15417,24970,38401,56640,80717,
%T A006529 
111762,151005,199776,259505,331722,418057,520240,640101,779570,940677,1125552,1336425
%N A006529 Cubes with sides of n colors.
%D A006529 M. Gardner, New Mathematical Diversions from Scientific American. 
Simon and Schuster, NY, 1966, p. 246.
%O A006529 0,3
%A A006529 njas
%K A006529 nonn

It's "b" sequence is 1 8 30 100 0 0 0 0...

as though there were 100 ways to 4-color the cube and no ways to
5 or 6-color it. What is Martin counting here?

Christian


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