[Guest]

0, 2, 8, 18, 7, 7, 9, 13, 10, 9, 1, 2, 5, . . .

If you can figure out the pattern, then let A=1, B=2, C=3, . . .

and find the first place in the sequence where nine consecutive letters

spell out two consecutive English words (letters are in order).

Without giving away the puzzle, include synonyms of both words somewhere in your response.

Edited December 13, 2009 by xamdam

[Guest]

Any hints?

[superprismatic]

I found a simple-to-generate cyclic 7920 pattern that contains this sequence but there are no words in it as you describe. In my generation scheme, the next 3 terms (after the 5) are 6, 26, 17. Perhaps you could supply more terms.

[Guest]

My function generates an infinite, non-cyclical sequence.

The first 21 elements are : 0, 2, 8, 18, 7, 7, 9, 13, 10, 9, 1, 2, 5, 10, 7, 7, 8, 11, 7, 4, 4, …

Both the minimum and maximum occur above.

While there are infinite occurrences of the maximum, the minimum occurs only once.

[Guest]

You might be able to find the key to the basic pattern by focusing on the first four elements.

[Guest]

You might be able to find the key to the basic pattern by focusing on the first four elements.

I was wondering earlier whether the first one or two or so elements were arbitrarily set, or if they were determined by the function.

Like, in the Fibonacci sequence (0, 1, 1, 2, 3, 5, 8...), the first two numbers are essentially random.

So I guess this answers that question

My biggest question would be this: is it a piecewise function (like, it undergoes a different equation if xn > 9 than does xn < 10), or is it the same kinda thing everywhere along the sequence?

Edited December 14, 2009 by DarthNoob

[Guest]

The closest thing I can get is this:

if an < 10

a(n+1) = 2a(n) + 2 - a(n-1)

and obviously something to handle cases where the number would go negative

But it doesn't work everywhere

Edited December 15, 2009 by DarthNoob

[Guest]

I was wondering earlier whether the first one or two or so elements were arbitrarily set, or if they were determined by the function.

Like, in the Fibonacci sequence (0, 1, 1, 2, 3, 5, 8...), the first two numbers are essentially random.

So I guess this answers that question

My biggest question would be this: is it a piecewise function (like, it undergoes a different equation if xn > 9 than does xn < 10), or is it the same kinda thing everywhere along the sequence?

Unlike the first two elements of the Fibonacci sequence, all elements are determined by the function.

It is not a piecewise function -- the sequence can be generated with one simple spreadsheet formula (containing no "if" statements or conditions on the domain).

However, the way the function behaves, it can appear to be in two (or even three) pieces:

0, 2, 8, 18, then 7, 7, 9, 13, 10, 9, then behaving nicely from here on 1, 2, 5, 10, 7, 7, 8, 11, 7, 4, 4, …

[Guest]

The closest thing I can get is this:

if an < 10

a(n+1) = 2a(n) + 2 - a(n-1)

and obviously something to handle cases where the number would go negative

But it doesn't work everywhere

If I'm understanding your equation right, a_(n+1) = 2a_(n) - a_(n-1) it does predict the 7th, 8th, 10th, 13th and 14th terms based on the previous term.

However, since my range ( or codomain?), unlike the Fibonacci sequence, is a finite set of integers (see previous hint re minimum and maximum) and the sequence is infinite I don't think it would be possible to write a formula that looks at one or two terms and predicts the next, would it? It seems like any finite sub-sequence you picked would eventually have to appear again and again in this infinite string.

How about if I tell you that this sequence can be generated if the domain of the function is the sequence of whole numbers (non-negative integers). (There is another domain that happens to work too.) The rule for the function is probably simplest to say in words (about 7 words), or to write as a spreadsheet formula.

[Guest]

"It seems like any finite sub-sequence you picked would eventually have to appear again and again in this infinite string. "

My bad. I was thinking of random numbers. You could easily design a sequence where certain sub-sequences don't repeat: 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, ... (here, 0, 1, 1, 0 won't repeat).

[Guest]

Guess I'll be the whistle-blower and say this one's not solvable; you couldn't crack it open with a machete.

Thanks though, Xamdam, for wasting my time!

[superprismatic]

each member of the sequence is the sum of the most significant digit of some number (which is 0 only at 0) with another digit from the same number. This number may be a simple function of a positive integer, but there are sooooo many of them!

Edited December 15, 2009 by superprismatic

[Guest]

does this have to do with the mod operator, for ex

512 mod 16 =0

512 mod 17 = 2

512 mod 18= 8

512 mod 19 = 18

However this function does not predict the numbers correctly after the first 4.

[Guest]

Guess I'll be the whistle-blower and say this one's not solvable; you couldn't crack it open with a machete.

Thanks though, Xamdam, for wasting my time!

And here I am working on your Twisted Decimal Code, 12 hours a day, staying home from work, 8 computers cranking away full-time, for a few weeks now,

and you quit in three days?!

[Guest]

each member of the sequence is the sum of the most significant digit of some number (which is 0 only at 0) with another digit from the same number. This number may be a simple function of a positive integer, but there are sooooo many of them!

Well done. You're on the mark so far. Most kids probably learn the fuction around 4th to 6th grade.

[Guest]

does this have to do with the mod operator, for ex

512 mod 16 =0

512 mod 17 = 2

512 mod 18= 8

512 mod 19 = 18

However this function does not predict the numbers correctly after the first 4.

No, it's a bit simpler than that.

That's a cool idea for your next puzzle post though.

(Check out superprismatic's reply and my response.)

Edited December 16, 2009 by xamdam

[Guest]

My function generates an infinite, non-cyclical sequence.

The first 21 elements are : 0, 2, 8, 18, 7, 7, 9, 13, 10, 9, 1, 2, 5, 10, 7, 7, 8, 11, 7, 4, 4, …

Both the minimum and maximum occur above.

While there are infinite occurrences of the maximum, the minimum occurs only once.

5, 8, 14?

Maybe it has to do with being right brained or left brained.

[Guest]

5, 8, 14?

Maybe it has to do with being right brained or left brained.

Firstly, those are indeed the 21st, 22nd and 23rd terms. Lucky guess?

And lastly, good luck finding the words. Are you using a spreadsheet?

Edited December 16, 2009 by xamdam

[DudleyDude]

I have to agree with ljb. Xamdam must be some kind of rat to try to stick us with this problem.

[Guest]

Firstly, those are indeed the 21st, 22nd and 23rd terms. Lucky guess?

And lastly, good luck finding the words. Are you using a spreadsheet?

I don't have a spreadsheet at home. I'll try that from work -- I'll definitely need one. (And no, it wasn't a lucky guess, you cruel person!)

Thinking up a new use for LJB's machete...

Edited December 16, 2009 by Patouie

[Guest]

I have to agree with ljb. Xamdam must be some kind of rat to try to stick us with this problem.

. . . . Whata buncha lightweights

[Guest]

I caught a glimpse of W's brother with some hot chicks, but in the end, I couldn't cut the mustard. Xamdam is definitely a weasel.

[Guest]

Kudos to ljb, DudleyDude and Patouie. Superprismatic gets credit for an assist. Very nice.

Edited December 17, 2009 by xamdam

[Guest]

Several questions unarg

e-------------------------------u

c-------------------------------a

n-------------------------------b

e--------------------------------l

u-------------------------------y

q--------------------------------

e-------------------------------a

s-----hcae gnidrager esir

[plainglazed]

I confess, cant cut it anymore, either.

[superprismatic]

Well, just call me 'Saber Skunk' from now on 'cause I misread the clue in post #9 to say 'positive' instead of 'non-negative' which really cut me to the quick!