[seqfan] Re: Question regarding A174283

[Rainer Rosenthal]

Dear Maximilian,

thanks to Hugo Pfoertner we now have the missing resistance. Its value
is 1/2.
So it is neither in the A048211(6) = 53 resistances
{1, 3, 6, 3/2, 2/3, 1/3, 4/3, 3/4, 5/6, 6/5, 9/2, 5/4, 4/5, 13/6, 13/7,
7/13, 11/6, 11/5, 5/11, 6/11,
9/7, 7/9, 9/4, 9/5, 5/9, 4/9, 10/3, 10/7, 7/10, 11/4, 11/7, 7/11, 4/11,
3/10, 12/5, 12/7, 7/12, 5/12,
11/3, 11/8, 8/11, 13/5, 13/8, 8/13, 5/13, 3/11, 2/9, 1/6, 6/13, 9/10,
10/9, 11/10, 10/11}
nor in the A174285(6) = 3 resistances {11/13, 1, 13/11}, thus completing
the A174283(6) = 57 resistances.

I suggest to show the 53 resistances as a not so trivial example a(6) in
addition to a(2) in A048211.
The example a(6) = 57 in A174283 can then easily given in short form as
A048211(6) + A174285(6) + 1.
The additional resistance 1/2 is the resistance of the following "6-arm"
bridge:
[code]
#
#
#                A-------.
#               / \      |
#             (1) (1)    |
#             /     \    |
#            B--(1)--C  (1)
#             \     /    |
#             (1) (1)    |
#               \ /      |
#                D-------'
#
[/code]

If you agree I will edit A174283 accordingly. And I will add those a(6)
= 53 resistances to A048211.

Cheers,
Rainer


Am 12.10.2020 um 16:59 schrieb M. F. Hasler:
> On Sun, Oct 11, 2020 at 5:28 PM Rainer Rosenthal <r.rosenthal at web.de> wrote:
>
> A174285 says "...confined to 5 arms...".
> So the additional one, counted in A174283(6), probably has more than 5
> arms...
>
> Also, the idea of  A174283 = A048211 + A174285 will go wrong (in the other
> sense)
> when the two distinct constructions yield common values, viz
> # ( A ∪ B ) = # A + # B  −  # ( A ∩ B ).
>
> -Maximilian
>
> --
> Seqfan Mailing list - http://list.seqfan.eu/