[seqfan] Re: Fwd: Worry about old sequence, A030077, paths in K_n

[Tim Peters]

[Brendan McKay]
> ...
> Thus, I believe all the values in the entry are correct.  The reason the
> graph looks dubious is that only for some values can distinct multisets
> have the same total.  For n <= 16, this happens only for 12 and 15.
>
> The number of distinct multisets of edge lengths up to n=16 is
> 1, 1, 1, 3, 5, 17, 28, 105, 161, 670, 1001, 4129, 6188, 26565, 38591,
> 167898,
> which is not in OEIS.

FWIW, my independent adventures in Python reproduced those results
through a(15) (the final results, the number of distinct chord
multisets, and that - through n=15 - at least one pair of distinct
multisets had the same total only at 12 and 15).

I used (although with care) plain old HW floating point to decide
whether two distinct chord multisets sum to the same length, but am
not nearly as frightened of that as group consensus appears to be. I
ran the PSLQ algorithm concurrently to look for integer relations
among {sin(j * pi / n}} for j in 1 .. floor(n/2), and there just
aren't any that get plausibly close for n as small as we're looking at
here.  For example, at n=14 PSLQ only found one candidate:

2*sin(5*%pi/14) + 2*sin(%pi/14) = 2*sin(3*%pi/14) + sin(7*%pi/14)

which is in fact an identity (BTW, multiply both sides by 2 if it
helps - I'm picturing a circle with unit diameter, not unit radius -
doesn't matter). Take any one of those terms out of the set PSLQ is
looking at, and it finds no distinct linear combinations that are
approximately equal.

FYI, it finds none for n=16 or n=17.

n divisible by 6 seems especially prone to this, because the identity

2*sin(%pi/6) = siin(%pi/2) = 1

arises then, which gives a very simple "building block" for creating
distinct multisets with equal sums (although for n=6 exactly, that
can't be done; n=12 can mix it with the simple identity

sin(%pi/12) + sin(3*%pi/12) = sin(5*%pi/12).
).

> Now I will run n=17.

But you'll have to try n=18 to confirm or refute the next instance of
my vague "divisible by 6" conjecture ;-)