# What's next in the sequence 1,6,16,32,56,89...?

What are the next terms in the sequence 1,6,16,32,56,89...?

Justify your answer! There could be more than one good answer...

[Jason] Hint: it isn't necessarily a continuous or differentiable function. Look at my recent question history.

[billblasphemy] You've gotta justify your answer...

[Ronatnyu] No it's not a Fibonacci sequence prank, and I wasn't thinking of that when I posted it, even if it looks similar. But having said that, you can use whatever method of analysis you want.

Vikram found the "correct" sequence I used, but Brian's is creative and whitesox's is pure ingenious, and works infinitely! I actually prefer whitesox's...

And Jason's comment was spot-on.

Actually Vikram's was slightly different;

My formula was simply ceil(x^2.5)

and using that, the next terms are 130,182,243 (not 245). In Python:

> [ int(ceil(x**2.5)) for x in range(21) ]

[0, 1, 6, 16, 32, 56, 89, 130, 182, 243, 317, 402, 499, 610, 734, 872, 1024, 1192, 1375, 1574, 1789]

I never noticed that √2π ~ 2.5 before...

### 10 Answers

- Vikram PLv 51 decade agoBest Answer
The next three terms will be 130, 182, 245

The generating function is f(n) = ceil(n^sqrt(2*pi)) - floor((n-1)/3)

check it out it will exactly give the sequence which you have given

Answer is based on your hint:

"[Jason] Hint: it isn't necessarily a continuous or differentiable function. Look at my recent question history."

So not much explanation is needed :)

Here goes first 50 terms

1, 6, 16, 32, 56, 89, 130, 182, 245, 319, 405, 505, 616, 743, 884, 1038, 1210, 1397, 1599, 1819, 2057, 2311, 2584, 2875, 3185, 3515, 3864, 4233, 4623, 5033, 5464, 5918, 6393, 6889, 7410, 7952, 8518, 9107, 9721, 10357, 11019, 11706, 12417, 13155, 13918, 14706, 15521, 16363, 17231, 18127 . . .

__________________________

I was trying to find "elegant" formula with mathematical constant, that's where I used π :-)

Interesting question BTW.

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Whitesox:

As given in the **hint** , I followed the related question which smci had asked

http://answers.yahoo.com/question/index?qid=200906...

Then have a look at the answer given by duke or the answer which smci gave in the related question referenced in the above question

http://answers.yahoo.com/question/index?qid=200905...

Hope this makes it clear :)

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@yo mismo:

Answer to your query:

If there are n terms, the polynomial of n-1 degree can fit all the values of sequence. There are many methods to generate this polynomial. The one which you have used is called Lagrange polynomial http://en.wikipedia.org/wiki/Lagrange_polynomial

which is based on order of differences between the terms and because of that if all terms in the given sequence are integer then the interpolated result will also give you an integer. (Read wiki for more information)

Other methods:

http://en.wikipedia.org/wiki/Bernstein_polynomial

http://en.wikipedia.org/wiki/Newton_polynomial

http://en.wikipedia.org/wiki/Polynomial_interpolat...

http://en.wikipedia.org/wiki/Hermite_interpolation

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@Yo Mismo:

It's easy to understand

x. . . . f(x)=x^2. .First order diff d(1) . . . . . . d(2)

1. . . ...1.. . . . . .. . . . . .3. . . . . .. . . . . .. . 2

2 . . . . 4 . . . . . .. . . . . .5. . . . . .. . . . . .. . 2

3 . . . . 9 . . . . . .. . . . . .7

4. . . . 16

Note that d(2) is constant so polynomial is of degree 2

Now if we want to "predict" (interpolate) next number in the sequence f(x) we can continue with d(2) = 2 & work in reverse order

x. . . . f(x)=x^2. .First order diff d(1) . . . . . . d(2)

1. . . ...1.. . . . . .. . . . . .3. . . . . .. . . . . .. . 2

2 . . . . 4 . . . . . .. . . . . .5. . . . . .. . . . . .. . 2

3 . . . . 9 . . . . . .. . . . . .7. . . . . .. . . . . .. . {2}

4. . . . 16. . . . . .. . . . . .{9=7+2}

5. . . . {25=16+9}

This should explain you why interpolated term is also an integer.

This when represented as polynomial will give you x^2

as shown in example 2 over here

http://en.wikipedia.org/wiki/Lagrange_polynomial

You can try it for other sequences using different f(x)

- Anonymous1 decade ago
I first thought about a polynomial which has degree <=5 because you gave 6 numbers.

P(x) = 3 - 439/60*x + 161/24*x^2 - 5/3*x^3 + 7/24*x^4 - 1/60*x^5

The next number would be P(7) = 129.

Edit:

It's not difficult to prove that if x is an integer then P(x) is also an integer.

But do you know if it's always like that?

I mean, you give us P(1)=p1, P(2)=p2,..., P(n+1)=p_(n+1)

with p1, p2,..., p_(n+1) integers.

Let P(x) the polynomial of degree <= n which passes through all P(i)=p_i you gave.

If x is an integer then P(x) is also an integer?

I know some of you here are really good at math and maybe you could help me, thanks!

Vikram: I think I need some time to understand why the interpolation in P(x) is always an integer. Your answer and links are very interesting, thank you.

Vikram: I can see it now, that was kind of you :-)

- whitesox09Lv 71 decade ago
I'm just going to find the next two terms, and this is probably a stretch, but here it goes...

Differences between each successive term:

5, 10, 16, 24, 33

Differences between the above differences:

5, 6, 8, 9

Which are the (Primes - 1)/2, beginning with 11.

Next in that sequence are 11, 14

So next in the differences sequence are 44, 58

So the sequence is:

1, 6, 16, 32, 56, 89, (89 + 44), (89 + 44 + 58), ...

Which is:

1, 6, 16, 32, 56, 89, 133, 191, ...

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Edit: Maybe I'm missing something here, but how did you figure out that formula Vikram?

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Vikram, I see now. I guess I just haven't been keeping up with the questions on this site lately.

- ?Lv 41 decade ago
I think the answer is 100.

1^2 = 1

6^2 = 36

16^2 = 256

32^2 = 1024

56^2 = 3136

89^2 = 7921

Then add the digits together:

1

9

13

7

13

19

Multiply the digits:

1

9

3

7

3

9

_

If you make it a number you get 193739_ (which I assume is a palindrome number)

100^2 = 10000, 1+0+0+0+0 = 1, and 1is a single digit so it stays alone and it would complete the palindrome number of 1937391.

So yeah, I think 100 is the next number in the sequence... >.>

Yeah, I totally bs'd that but at least I tried. :(

~♫♪♫

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- ?Lv 51 decade ago
I'll admit I can't crack it. I'm annoyed by the sudden appearance of an odd number. Also the (apparently coincidental) doubling from 16 to 32, where both are already powers of 2.

Good luck all. =)

- billblasphemyLv 61 decade ago
119. the sequence has started with 5 and doubled each time.

Source(s): genius - RonatnyuLv 71 decade ago
Chickens are after my real false teeth

Source(s): ..... is this one of those Fibonacci sequence pranks? - Anonymous1 decade ago
*brainsplosion*

Source(s): I only starred because I'm too stupid to figure it out and need to know. I didn't realize all of these rascals would follow.