Still More Fun, Easy Sequences

[cubestudent]

i love sequence puzzles.

here are two more i thought were pretty cool (though perhaps not too tough).

S1:
1, 3, 8, 20, 48, 112, 256, ?, . . .

S2:
1, 2, 5, 15, 52, 203, ?, . . .


have at them!

[edit]ouch! i typoed in S2. sub'd 22 for 2. now it's fixed[/edit]

[This message has been edited by cubestudent (edited 04-01-2002 02:19 AM).]

[Zealot]

I've never really tried any puzzles like this before, but I think I have the first one. The first term subtracted from the second, multiplied by 4, gives the next term.

[Zealot]

Of course, I forgot to post the next number in the sequence...
Which should be 576.

[cubestudent]

dead on.


your answer: S(x) = 4 * (S(x-1) - S(x-2))
one of the things i love most about these puzzles is the different viewpoints.

my answer was: S(x) = S(x-1) * (2 + 2/(x))
neato, eh?


[edited to fix stupid typo]

[This message has been edited by cubestudent (edited 04-01-2002 02:38 AM).]

[Zealot]

Give me about 20 minutes and I might understand what you just said...

[cubestudent]

x refers to the position of a term.
eg. in S1, S(1) = 1; S(2) = 3; S(3) = 8; . . .

so S(x) is the next term we're calculating.
your answer uses S(x-1) (the previous term), and S(x-2) (the term two spots back).

example:

code:

---

looking for term 5:


S(x) = 4 * (S(x-1) - S(x-2))


S(5) = 4 * ( S(4)  -  S(3) )


S(5) = 4 * (  20   -   8   )


S(5) = 4 * (       12      )


S(5) = 48

---

[This message has been edited by cubestudent (edited 04-01-2002 04:51 AM).]

[Zealot]

I got that much, but this S(x) = S(x-1) * (2 + 2/(x+1)) is confusing me. I tried substituting the first three terms in, but couldn't see how it worked.

[cubestudent]

aaack!

you couldn't get it because i typoed again!

so sorry!

the (x+1) was just supposed to be x
i got my indexes mixed up

[Zealot]

code:

---

S(x) = S(x-1) * (2 + 2/(x))


S(8) = (3) * (2 + 2/(8))


S(8) = (3) * (9/4)


S(8) = (27/4)

---

??

[cubestudent]

code:

---

S(3) = S(2) * (2 + 2/3)


S(3) =  3   *  (2 2/3)


S(3) = 8





S(8) = S(7) * (2 + 2/8)


S(8) = 256  *  (2 1/4)


S(8) = 576

---

[This message has been edited by cubestudent (edited 04-01-2002 02:55 AM).]

[Jingle47]

I am trying to work on S2. ARe the numbers correct for sure now?

[Zealot]

Gotcha. I was using S(x) for the denominator at the end, not x.

Thanks.

[cubestudent]

Jingle:
the numbers for S2 are definitely: 1, 2, 5, 15, 52, 203, ...

[Zealot]

My first guess would be 809. Am I way off?

[cubestudent]

that really depends on your reasoning
it's certainly not the answer i was looking for (but you are w/in 100 ).

[Jingle47]

I got the same answer as zealot S(2) =S(1)x4-2
S(3)=S(2)X4-3
S(4)=S(3)X4-5
S(5)=S(4)X4-8
S(6)=S(5)X4-5
SO S(7) should be(according to our twisted logic) s(6)x4-3=809.

------------------
I AM THE SMARTEST GUY ON THE BLOCK!
-Psst! "He lives in a cul-de-sac"

[cubestudent]

hmm. i dunno:

if the pattern went: *4-2, *4-3, *4-5, *4-8, *4-13, ... (following Fibonacci's)

i would hail your answer as marvelous, and a far more elegant alternate solution than was mine.

but because it goes: 2, 3, 5, 8, 5, 3. . .
i'd say it's an answer, but not a very good one.

i realize, of course, that my only giving six terms is a lot of the problem. i'll add a seventh in a bit, if no one comes up with a good answer.



[This message has been edited by cubestudent (edited 04-01-2002 06:20 AM).]

[Zealot]

Well, yes, that is the twisted logic I was using , but that sequence doesn't continue very far, so I thought it unlikely to be the answer.

[cubestudent]

upon further consideration, i would like to label S2 as a Very Hard Sequence Puzzle (TM).

my rating system runs something like:

code:

---

1) Easy - even those unfamiliar with sequences can get these.


2) Normal - requires work/thought/insight for sequence newbies,


            or a simple systematic approach for veterans


3) Hard - requires a fair amount of work/thought and/or an Aha! moment.


4) Very Hard - requires a lot of work/thought and an Aha! moment.


5) Nuts! - requires idiot savancy, or


           a very lucky series of ideas combined with a lot of work/thought


 


as examples:


1, 2, 4, 7, 11, 16, 22, ... is an Easy Sequence Puzzle


  ( add 1, then 2, then 3, ... )


  ( aka. S(x) = S(x-1) + (x-1) )


1, 1, 2, 3, 5, 8, 13, ... is on the line between Easy and Normal (maybe 1.5)


  ( Fibonacci's: add the two previous terms )


  ( aka. S(x) = S(x-1) + S(x-2) )


my S1: 1, 3, 8, 20, 48, 112, 256, ... is Hardish Normal (perhaps a 2.3 )


  ( as answered by Zealot: 4 times the difference of last two terms )


  ( aka. S(x) = 4 * (S(x-1) - S(x-2)) )


Chuck's original Easy, Fun (find this yourselves) is Very-Hardish Hard (~3.7)

---

i think my S2 is a solid 4.
so i don't think it will hurt to give the seventh and eighth terms:

S2: 1, 2, 5, 15, 52, 203, 877, 4140, ...

enjoy!



[This message has been edited by cubestudent (edited 04-01-2002 06:18 AM).]

[groza528]

Originally posted by cubestudent:

---

if the pattern went: *4-2, *4-3, *4-5, *4-8, *4-13, ... (following avogadro's)

---

Would that be Avogadro the chemist, who found the number for the mole? Or Avogadro the mathematician whose real name is Fibonacci?

[cubestudent]

aack!

i. . .

double aack!

*turning very red*

actually, , i have no idea what you're talking about.

i love editing.

[This message has been edited by cubestudent (edited 04-01-2002 06:19 AM).]

[mith]

The sequence has something to do with e. I won't post the next term because I am too lazy to figure it out.

[edit]Actually, I'm not. The next is 21147. I don't know if *your* reasoning has to do with e, but mine does.

[This message has been edited by mith (edited 04-01-2002 09:48 AM).]

[cubestudent]

wow.

you are, of course, correct.
(perhaps a 4 was the wrong rating after all)

i must know how you got this.
my rule is clumsy and ridiculous and has nothing to do with e.

[mith]

These are called Bell Numbers. There's several different ways to get them (I was reading about them earlier after I posted the answer). The one I knew (I discovered them in high school, before I knew what they were called) is:

1/1!+2/2!+3/3!+... = e
1/1!+4/2!+9/3!+... = 2e
1/1!+8/2!+27/3!+...= 5e
etc.

The Bell Numbers also give the number of ways a set of points can be partitioned. They can also be calculated using a Bell Triangle:

code:

---

1


1  2


2  3  5


5  7  10 15


15 20 27 37 52

---

[cubestudent]

i used the bell triangle to find the numbers, myself

i'm still a little unclear on the relation to e, though. i still don't see any connection between the two.

if this is slow of me, please show some forebearance, i've been out of the mathematics game for the last six years, minus the last three weeks.

[mith]

This part:

quote:

---

1/1!+2/2!+3/3!+... = e
1/1!+4/2!+9/3!+... = 2e
1/1!+8/2!+27/3!+...= 5e

---

In general, B(n) = (1/e)*sum(k=1 to infinity) of k^n/k!

(i'm too lazy to figure out symbol font )

There's some other cool relations too; just search google.

[mith]

The partition bit is probably the most important one, so I will give some examples for anyone reading this that is too lazy to go searching.

Partitions of {1} : {1}
Partitions of {1,2} : {1,2}, {1|2}
Partitions of {1,2,3} : {1,2,3}, {1,2|3}, {1,3|2}, {2,3|1}, {1|2|3}

etc.

[cubestudent]

oh! *smacks forehead* the coefficients!

e, 2e, 5e didn't immediately register as the beginning of 1, 2, 5, 15, ...

thanks for the info.

got any other sequences for us to enjoy?

[CrystyB]

So there IS a way to compute the # of partitions! Yay!

[cubestudent]

Crysty:
and to make it a little easier,
this is the tree of differences

code:

---

              52


           15    67


         5    20    87


      2     7    27   114


   1     3    10    37    151


1     2     5    15    52    203

---

note the equivalence between the two 'lines' of bolded elements.
to continue the tree, just add the newest term (here 203) to the top of the tree.

[CrystyB]

you're telepatic!!! I NEVER asked HOW to do it. I was just happy i have new info. BUT it seems that after that my brain decided to post this:

"CubeStudent, want a third viewpoint on S1? S(n+1)=2*S(n)+2^n if you number them from 0 or +2^(n-1) if from 1.

Also, you said you want a seq? Try this one! Though i'm probably being mean...

Also, care to explain how you STARTED the triangle? I mean you compute a partial one and put the number found (lower-right corner) as first on next line and then start to fill that line, and so on? ((it certainly appears that way!))"

Now how could you have foreseen that???

[This message has been edited by CrystyB (edited 04-13-2002 07:32 PM).]

[cubestudent]

foreseen what?
i'm completely confuzzled.

[CrystyB]

i posted that "TY mith" post (my first and only post in here until yesterday) AND NEVER opened this topic again. Till yesterday that is. But on further inspection i became puzzled and composed (offline) the QUOTED part of the above post, only to discover when i opened this to "post reply" that you have answered my question 12 days before i posted it!!!