[seqfan] An idea for a simple mechanical computer to illustrate two OEIS-sequences.

[Antti Karttunen]

A correction to the question (1). It should be:

(1) How many bean-moves have been made from prime cups before the
prime cups are empty?  Note that this count includes moves between the
prime cups, and also all the moves from cup 2 to the reservoir-cup.
However, this does NOT include the moves from the reservoir-cup to the
prime cups.

And there's a slight modification to that, question (1b): How many
bean-moves have been made from the prime cups before the all prime
cups with an odd prime label are empty? That is, now we stop the
solitaire immediately when we see that only the cup 2 has any beans
left. Also, we could ask how many beans will be in the cup 2 at that
point. I think also this one is in OEIS: 0, 1, 1, 2, 2, 1, 1, 3, 1, 2,
2, 2, 2, 1, 3, 4, 4, 1, 1, 3, 2, 2, 2, 3, ...
Well, not..., but instead the sequence giving one larger counts is there.


Best regards,

Antti



On 8/28/21, Antti Karttunen <antti.karttunen at gmail.com> wrote:
> Seems that people have had difficulties in visualizing my machine.
> So let's try this one instead:
>
> Take a mancala, wari, etc. board (see e.g.,
> https://en.wikipedia.org/wiki/Oware )
> and designate one of the cups as "reservoir", and put all your beans
> there first.
> Label the other cups with successive primes, "2",  "3", "5", "7",
> "11", and so on.
> Initialize the "computer" (or "solitaire") by transferring from the
> reservoir-cup zero or more beans to each cup. The starting value of n
> is given by the product of those primes, with the number of beans in
> each giving the exponent of the corresponding prime.
>
> Then start the solitaire, by making "rounds" over the "prime-cups",
> from the one labeled "2" to the highest prime, in each round doing:
>
>   if there are any beans in cup 2, move one bean from the cup 2 back
> to the reservoir, and if there are not, then just skip it, to the cup
> 3,
>
>  next move one bean from the cup 3 to the cup 2 (similar proviso here,
> if no beans in cup 3, then just skip it),
>
>  next if there are any beans in the cup 5, move one bean from it to
> cup 2, AND also one bean from the reservoir to cup 2 (so the beans in
> cup 2 got incremented by 2),
>
>  next if there are any beans in the cup 7, move one bean from it to
> cup 2, AND also one bean from the reservoir to cup 3, etc.
>
> That is, in general, if there are any beans in the cup labeled "p" (p
> an odd prime), move one of them to the cup "2", and take from the
> reservoir enough beans to fill the cups according to the prime
> factorization of (p-1)/2.
>
> The questions asked are:
>
> (0) Does the solitaire eventually end, with all beans back in
> reservoir? (Yes, it's easy to see that, so not really a question for
> this forum.)
>
> (1) How many times a bean has been put back in the reservoir when the
> solitaire is over? (Note that this just counts the times the player
> has transferred a bean from the "prime-cups" to the "reservoir-cup".
> The beans are indistinguishable from each other).
>
> (2) How many such rounds over the prime-cups will be made before they
> are empty, and all beans are back in the reservoir? (I give a hint to
> this at the bottom of the page, just scroll down.)
>
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> Hint to the question 2: A000010(n) = A003557(n) * A173557(n).
>
> And the question 1 relates to a problem that Bob Wilson and several
> other people were pondering about in the Spring 2020.
>
>
> Best regards,
>
> Antti
>
>
> On 8/26/21, Antti Karttunen <antti.karttunen at gmail.com> wrote:
>> I went to pick up some berries, but came back with this idea instead
>> (well, in any case, involving small round things):
>>
>> We have a machine, with several glass tubes, that are labeled as
>> "2", "3", "5", "7", "11", "13", etc. When machine is in
>> operation, each tube contains zero or more steel balls that just
>> fit inside them.  The tubes are open from the top, while at the bottom
>> of each one there is a mechanical "valve", normally closed, but which
>> can be electrically opened just for the time it takes for exactly one
>> ball to come out of the tube (or none if the tube is empty, but never
>> letting out more than one).  Moreover, at the top of the glass tubes is a
>> moving "ball-dispenser" that may automatically drop one or more balls to
>> some
>> of the glass tubes, by the program's instructions.  (I'm partly
>> inspired by this old Lotto-machine https://youtu.be/_5Dn7mm2WkE?t=21
>> but of course in our version the tubes must be tall enough to
>> accommodate several balls, balls that however, need not be specially
>> marked, but can be totally indistinguishable from each other).
>>
>> Also, under each bottom-valve of each tube is either an optical or
>> mechanical sensor that detects a dropped ball, before it is transferred
>> back up to the dispenser's reservoir of extra balls by some mechanism
>> hidden to us.
>>
>> The user activates the machine by manually dropping several balls into
>> the tubes (>= 0 in each), after which he presses the start button,
>> from which onward the machine's mechanical program proceeds as follows:
>> the
>> tubes are scanned from the left to the right, starting, from the tube
>> labeled "2", with each tube's bottom valve opened in turn to allow a
>> single ball to come out of the tube, provided there are any balls in that
>> tube.
>> The dropped ball then passes the above-mentioned "ball-sensor", which
>> activates a hard-coded "dispensing routine" specific for each tube.
>>
>> For the tube "2" nothing is done, for the tube "3", for any ball that
>> comes from the bottom of it, the ball-dispenser drops an extra ball
>> into the tube "2" (*), for the tube "5", if a ball comes out, then the
>> ball-dispenser drops two balls into the tube "2", for the tube "7",
>> for a single ball coming out, the dispenser drops one ball to "2" and one
>> ball to "3", for "11" one ball to "2" and one to "5", for "13" two
>> extra balls to "2" and one to "3", and so on.  Each ball-sensor,
>> when activated by a dropped ball, also rotates an arrow in a clock-like
>> display, which keeps the total count how many balls goes through the
>> machine.
>> Additionally, there is another arrow that is rotated by one step
>> whenever all the tubes have been scanned in turn and the process
>> returns back to the tube "2". The output of the program are those two
>> counts, which are ready when all the tubes are finally empty.
>>
>> (* Here it might be the same ball, if there were a direct intermediate
>> tube or channel
>> from the bottom of "3" to the top of "2". For other tubes no such
>> shortcut-optimizations
>> are possible, and the dispenser needs to be used.)
>>
>> If we consider the number of balls in the tubes "2", "3", "5", ..., etc,
>> as giving the exponents in the prime factorization of n, which two
>> sequences a(n) and b(n) the counts given by those two arrows correspond
>> to?
>> (Both are in OEIS).
>>
>> I don't know whether this kind of machine would make any sense in a
>> science fair (the mathematics doesn't involve Fibonaccis, but just basic
>> number theory, no rocket science), but it might make a nice clanking
>> contraption for Bridges kind of Art festival? Or let the bright kid to
>> play with it and figure out what's going on? Like e.g., why the machine
>> always stops, even though for a moment the number of balls often
>> increases?
>>
>> Also, I wonder if there there other number theoretical iterations or
>> problems that would allow such simple-minded mechanical
>> implementations?
>>
>>
>> Best wishes,
>>
>> Antti Karttunen
>>
>