[seqfan] Re: The Savannah Math

[Ali Sada]

 Hi David,
Thank you very much for your response. As always, I really appreciate it.
I excluded the empty sets because including them wouldn't give me a unique sequence. I didn't know how to word the definition in a way that makes this clear.

As for the hungry lions, H+Z=F. A hungry lion eats one Zebra to become a fed lion. 

For example, let's take the combination F3Z. We have a fed lion with three zebras. In the next week, the fed lion becomes a hungry lion, eats one of the zebras, and becomes a fed lion. The resulting combination is F2Z. The next week it becomes FZ, then it becomes F, then H, then it dies. 

Best,
Ali

    On Thursday, December 26, 2019, 8:19:02 AM EST, David Seal <david.j.seal at gwynmop.com> wrote:  
 
 > 2.    In the second week, the fed lion becomes a hungry lion, and the hungry lion dies. So, we have only two possibilities left (H or Z). When another unknown animal (F or H or Z) comes, we have the following combinations: 
> F+H=FH
> F+Z=FZ
> H+H=2H
> H+Z=F
> Z+Z=2Z
> We have 5 distinct possibilities; so a(2)=5

I don't follow. We have three possibilities left before the new animal arrives, namely H or Z or - (no animals left). Those combine with the three possibilities for the new arrival as follows:

H+F = FH
H+H = 2H
H+Z = HZ
Z+F = FZ
Z+H = HZ
Z+Z = 2Z
-+F = F
-+H = H
-+Z = Z

I make that 8 possibilities: (F, H, Z, FH, FZ, 2H, HZ and 2Z), with HZ arising either from a fed lion arriving in week 1 and a zebra in week 2, or from a zebra arriving in week 1 and a hungry lion in week 2.

I'm not clear about what HZ then becomes in week 3, before the new animal arrives. I can see three possibilities:

* The hungry lion dies, already being too weak to catch and eat the zebra, and the zebra survives - so we're left with Z.

* The hungry lion catches and eats the zebra, becoming a still-hungry lion and leaving us with H. This basically corresponds to saying that a lion requires one zebra every week - it can miss out one week but a second week will kill it off.

* The hungry lion catches and eats the zebra, becoming a fed lion and leaving us with F. This basically corresponds to saying that a lion requires one zebra every two weeks - though it will eat a zebra every week if it can (judging by your statement that FZ becomes Z over a week - though your stated rules don't actually state that a fed lion eats a zebra).

Under the last two possibilities, there's also the issue at some point of what FHZ becomes: if the hungry lion catches and eats the zebra, the fed lion becomes hungry and we're left with FH or 2H (depending on what HZ becomes); if the fed lion catches and eats the zebra, the hungry lion dies and we're left with F.

To summarise, I think you need to include the possibility of having an empty set of animals left in your counting, and you need to clarify your rules about when lions eat zebras and possibly also about *which* lions eat zebras.

David


> On 21 December 2019 at 22:05 Ali Sada via SeqFan <seqfan at list.seqfan.eu> wrote:
> 
> 
> Hi Everyone,
> 
> The setup below has no real meaning in zoology. It is just a math practice. The math system is based on the following assumptions:
> a)    One hungry lion (H) eats one zebra (Z) to become a fed lion (F).
> b)    In one week, a fed lion becomes hungry; a hungry lion dies; and zebras don’t change.
> c)    Every week, a new unknown animal (either F, H, or Z) enters the area. 
> 
> Let’s count the distinct possibilities for each week:
> 
> 1.    In the first week, we have one unknown animal, so, have 3 possibilities: it could be either F, H, or Z.  So, a(1)=3
> 
> 2.    In the second week, the fed lion becomes a hungry lion, and the hungry lion dies. So, we have only two possibilities left (H or Z). When another unknown animal (F or H or Z) comes, we have the following combinations: 
> F+H=FH
> F+Z=FZ
> H+H=2H
> H+Z=F
> Z+Z=2Z
> We have 5 distinct possibilities; so a(2)=5
> 
> 3.    In the third week, the combinations of the second week change:
> FH becomes H; FZ becomes F; 2H disappears; F becomes H; and 2Z remains 2Z.
> So, we have only 3 distinct possibilities (F, H, and 2Z) 
> When a new animal comes, the distinct possibilities become: (F, 2F, FH, FZ, F2Z, 2H, 3Z). 
> So, a(3)=7. 
> If we continue, we get a(4)=11, a(5)=14, a(6)=17, a(7)=22, etc.
> 
> I would really appreciate it if someone could help me find the terms of this sequence. Also, what’s the pattern here?
> 
> Best,
> 
> Ali
> 
> 
> 
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