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From: Leroy Quet on 1 Jun 2010 16:48
Here is a game for two players.

The players take turns picking any integer from 1,2,3,...., r that has
not been picked (by either player) previously in the game, where r is
some large integer (such as 1000 or 10000 or more).

Let this picked number be m. (m was picked by the player temporarily
called the "provider".)
The same player then picks any integer k where 1 <= k <= m. (k may
have been picked earlier any number of times in the game.)

The other player (the "finder") then tries to find any positive
integer n not equal to m (n can be arbitrarily large and either picked
previously during the game or not) such that:

Both d(m) = d(n) and d(m+k) = d(n+k), where d(j) is the number of
divisors of j.

The finder may also provide a proof that no such n (not equal to m)
exists.

If the finder either finds an n or proves there is no such n fitting
the conditions, then the finder gets a point.

Players then switch who is the provider and who is the finder.

Players play an even predetermined number of moves, and the player
with the largest score wins.

Thanks,
Leroy Quet

From: Leroy Quet on 2 Jun 2010 09:59
Hmmm. This game definitely doesn't require a strategy, in the
traditional sense. Unless you count "be better than your opponent at
math" as a strategy. It definitely is still a mental game, though,
obviously.

Note: For those of you who don't know, some ordered pairs (d(m), d(m
+k)) only occur once among (d(j), d(j+k)), j >= 1. The trick if you
are the provider is to find such an ordered pair that is difficult for
your opponent to prove is unique.

Thanks,
Leroy Quet

From: Skarfy on 3 Jun 2010 14:53
This does not seem like an enjoyable game. Have you ever played it
with anyone? Or is this simply to discuss math theory?


On Jun 2, 7:59 am, Leroy Quet <qqq...(a)mindspring.com> wrote:
> Hmmm.  This game definitely doesn't require a strategy, in the
> traditional sense. Unless you count "be better than your opponent at
> math" as a strategy. It definitely is still a mental game, though,
> obviously.
>
> Note: For those of you who don't know, some ordered pairs (d(m), d(m
> +k)) only occur once among (d(j), d(j+k)), j >= 1. The trick if you
> are the provider is to find such an ordered pair that is difficult for
> your opponent to prove is unique.
>
> Thanks,
> Leroy Quet

From: Chip Eastham on 3 Jun 2010 19:28
On Jun 2, 9:59 am, Leroy Quet <qqq...(a)mindspring.com> wrote:
> Hmmm.  This game definitely doesn't require a strategy, in the
> traditional sense. Unless you count "be better than your opponent at
> math" as a strategy. It definitely is still a mental game, though,
> obviously.
>
> Note: For those of you who don't know, some ordered pairs (d(m), d(m
> +k)) only occur once among (d(j), d(j+k)), j >= 1. The trick if you
> are the provider is to find such an ordered pair that is difficult for
> your opponent to prove is unique.
>
> Thanks,
> Leroy Quet

I'm one of those who don't know. Can you
show me an example of proving uniqueness
of a pair d(m),d(m+k)? Obviously d(1) = 1
is a special case, but I haven't thought of
another "angle" to prove uniqueness.

regards, chip
From: Leroy Quet on 4 Jun 2010 10:38


Chip Eastham wrote:
> On Jun 2, 9:59 am, Leroy Quet <qqq...(a)mindspring.com> wrote:
> > Hmmm.  This game definitely doesn't require a strategy, in the
> > traditional sense. Unless you count "be better than your opponent at
> > math" as a strategy. It definitely is still a mental game, though,
> > obviously.
> >
> > Note: For those of you who don't know, some ordered pairs (d(m), d(m
> > +k)) only occur once among (d(j), d(j+k)), j >= 1. The trick if you
> > are the provider is to find such an ordered pair that is difficult for
> > your opponent to prove is unique.
> >
> > Thanks,
> > Leroy Quet
>
> I'm one of those who don't know. Can you
> show me an example of proving uniqueness
> of a pair d(m),d(m+k)? Obviously d(1) = 1
> is a special case, but I haven't thought of
> another "angle" to prove uniqueness.
>
> regards, chip

An easy example, d(3) = 2, d(4) = 3. For all pairs (d(m),d(m+1)), m =
3 is the only example. (Because p^2-1 is composite for all p >= 3.)
There are many more examples, possibly an infinite number.

Thanks,
Leroy Quet

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