Number Sequences

People are often posting number sequences to this group asking what number comes next in the sequence.

Of course, the technically correct answer is that there are an infinite number of numbers that could come next.

The person posting the question might then respond that they are looking for the simplest possible answer.

Well, here is an example where the simplest possible answer isn't what you might first expect.

What digit comes next in the following number?

3.141592?

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Most people would immediately say '6' thinking immediately of the decimal expansion of pi:

3.1415926535897932384626433832795...

However, pi is the limiting result of an infinitely long formula.

That certainly isn't a very simple answer!

There is a much simpler equation that generates the given number:

355 / 113

Therefore, the number that actually comes next is '9' and not '6'.

The number actually continues:

3.1415929203539823008849557522124...

It just looks like pi if you consider only the first 7 digits!

Depends how you define 'simpler' surely. 355/113 is rational as opposed to pi being irrational, but pi is a number that arises a lot more - one could say it's 'more famous' than 355/133 which is just the ratio of a fairly random pair of integers which happens to be close to pi.

regards

> comes next in the sequence. > > The person posting the question might then respond that they are looking for > the simplest possible answer. > > Well, here is an example where the simplest possible answer isn't what you > might first expect. What digit comes next in the following number? > > 3.141592?

Now THAT is very cute indeed! Best remark we've had on this thread for a long time, and best post of the week! I intend to steal it ruthlessly!!

Meanwhile...

A third possible answer is 7. Since we're only being asked for one more digit, then 'obviously' there is only one more. The number appears to be an approximation of pi, and the best approximation of pi with 7 decimal places is 3.1415927 (since the last digit rounds up). QED.

'Gary Edstrom' wrote ... [...] What digit comes next in the following number? 3.141592? [...] Most people would immediately say '6' thinking immediately of the decimal expansion of pi [...] There is a much simpler equation that generates the given number: 355 / 113 [...] The number actually continues: 3.1415929203539823008849557522124...

That should be '...123...' instead of '...124...'. 355 / 113 has a repeating decimal segment of 112 digits:

3.14159292035398230088495575221238938053097345132743362 831 8584070796460176991150442477876106194690265486725663716 8 1415929203539823008849557522123893805309734513274336283 1 8584070796460176991150442477876106194690265486725663716 8 ...

ObPuzzle: Other than direct calculation, is there a formula that gives f(p,q) = length of the shortest repeating decimal segment in rational fraction p/q?

(E.g., f(355, 113) = 112, f(2, 4) = 1, f(8, 7) = 6, etc.)

What would a 3-dim plot of f(p,q) look like?

Perhaps 3 as the answer to 31415920/9999999 or 1 as the answer to 3141589/999999 or 4 as the answer to 3141561/999990 . . . 2 as the answer to 2827433/900000

ie as various repeating decimals expressed as fractions.

Regards John Collins

Since you include f(2,4)=0.50000000........ you are including repeating zeroes.

It is independent of p and only depends on q assuming p/q is reduced to its simplest form (p and q have no common factors).

If q factorises then remove any powers of 2 or 5 to get q1.

If q1=1 then you have a repeating zeroes fraction so the answer is 1.

For any other value of q1 the answer is the length of the shortest integer consisting of 9s only for which q1 is a factor. At the moment I cannot see a method other than direct calculation to do this.

Regards John Collins

'Dr J D Collins' wrote ... ['r.e.s.' wrote ...] > ObPuzzle: > Other than direct calculation, is there a formula that gives > f(p,q) = length of the shortest repeating decimal segment in > rational fraction p/q? > > (E.g., f(355, 113) = 112, f(2, 4) = 1, f(8, 7) = 6, etc.) > > What would a 3-dim plot of f(p,q) look like? > Since you include f(2,4)=0.50000000........ you are including repeating zeroes.

Yes. (Or repeating nines, since f(2, 4) = 0.5000... = 0.49999..., etc.)

It is independent of p and only depends on q assuming p/q is reduced to its simplest form (p and q have no common factors).

If q factorises then remove any powers of 2 or 5 to get q1.

If q1=1 then you have a repeating zeroes fraction so the answer is 1.

For any other value of q1 the answer is the length of the shortest integer consisting of 9s only for which q1 is a factor. At the moment I cannot see a method other than direct calculation to do this.

Ok, that follows from the fact that if 1/q = 0.RRR..., then (0.RRR...) * (999...9) = (0.RRR...) * (10^k - 1) = R where R is the repeating decimal segment written as an integer and the number of nines is k = f(1, q).

So that's a viscious circle, isn't it? (You'll need to find f(1 , q) in order to find the number of nines in order to find f(1, q) ...).

[...] It is independent of p and only depends on q assuming p/q is reduced to its simplest form (p and q have no common factors).

If q factorises then remove any powers of 2 or 5 to get q1.

If q1=1 then you have a repeating zeroes fraction so the answer is 1.

For any other value of q1 the answer is the length of the shortest integer consisting of 9s only for which q1 is a factor. At the moment I cannot see a method other than direct calculation to do this.

Ok, that follows from the fact that if 1/q = 0.RRR..., then (0.RRR...) * (999...9) = (0.RRR...) * (10^k - 1) = R where R is the repeating decimal segment written as an integer and the number of nines is k = f(1, q).

So that's a viscious circle, isn't it? (You'll need to find f(1 , q) in order to find the number of nines in order to find f(1, q) ...).

Oops, sorry. I should have realized that you were referring to simply stepping through the direct calculation of checking whether q divides 999...9, with the number of nines increasing by one each time until it does.

That's a good point. What is the simplest solution?

Douglas Hofstadter tells the story of when he was discussing AI and pattern matching and Richard Fenyman kept interjecting that the next number in the sequence is always 4, much like Richard Heathfield answered 42 to the 'b. NAME IT ?' thread.

Hofstadter accused Feynman of playing the 'village idiot'. Of course patterns like the ones he presented must seem entirely arbitrary to a computer, but Hofstadter was trying to discuss machine intelligence. If a machine was programmed to think like a human then what patterns would it see?

I wasn't able to solve the aforementioned puzzle, but I was at least able to see a pattern (by examining the derivatives):

0?, 65, -1

593,

The first order derivative is monotomically decreasing and the second order derivative is bounded between -1 and -5. These are quite clearly patterns that are recognizable to humans. Whatever answer I give should at least obey these two rules.

If I flip a coin 5 times and it lands heads (or tails) 5 times in a row, most people would see a pattern. Of course, the probability of this happening is only 1/16, but most people would still see a pattern.

Aren't all these puzzles intended for the entertainment of humans? After all, if I give you a crossword to solve, you can quite reasonably claim that on the planet Zarquox, '17a.. Indiscreet' is 'zcvdfgw', but the implication is that the answers must be supplied in English, as governed by common usage, and not according to your own made-up dialect.

The Richards of this world are much misunderstood, despite our natural qualities of genius, perfectionism, and (of course) modesty. We suffer in silence. Did you know Einstein was actually christened Richard Albert Einstein? True. [1]



<snip>

[1] Um... or possibly false. One of those.