[seqfan] Re: Question regarding A174283

[Rainer Rosenthal]

Dear Maximilian,

sorry, I did not get the meaning of "(in the other sense)" immediately.
Now I got it: that was just a general remark how equalities may "go wrong".

The interesting part of your answer was this one:
M. F. Hasler:
So the additional one, counted in A174283(6), probably has more than 5
arms...

But no, that is not the case since A174283 says "confined to the five
arms ...".

The good thing is that n = 6 is exemplified in A174283:
Example-2: 6 equal resistors, four arms have one unit resistor each and
the fifth arm has two unit resistors. Two resistors in the same arm,
when combined in series and parallel result in 2 and 1/2 respectively
(corresponding to 2: {1/2, 2} inA048211 <https://oeis.org/A048211>). The
set {1/2, 2}, in the diagonal results in {1}. Set {1/2, 2} in any of the
four arms results in {11/13, 13/11}. Consequently, with 6 equal
resistors, we have the set {11/13, 1, 13/11}, whose order is 3.

No mentioning of more than five arms! The example is a bit wanting in
that it tells you only about 3 resistance networks. The number a(6) = 57
is not mentioned explicitly. It is left to the reader to find out where
the other 54 resistance networks come from.
The title of A174283 is: "Number of distinct resistances that can be
produced using n equal resistors in series and/or parallel, confined to
the five arms (four arms and the diagonal) of a bridge configuration.
Since the bridge requires a minimum of five resistors, the first four
terms are zero" and the 3 resistance networks in the example are bridges
with 5 arms. From that I deduce that I have to look for 54 bridge-less
resistance networks with 6 resistances. But there are only A048211(6) =
53 of these.

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:
>
>> A048211 : The terms of this sequence consider only series and parallel
>> combinations; A174283 considers bridge combinations as well. - Jon E.
>> Schoenfield, Sep 02 2013
>> ~~~~~~~~~~~~
>> For bridge combinations there is A174285 = 0, 0, 0, 0, 1, 3, ...
>> So I expect A174283 to be the sum, but  A174283 = 1, 2, 4, 9, 23, 57, ...
>>
> 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
>
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