Math for fun, 1, 2, 3, 4, 5, 8, ____, ____, calculus 2 'find the pattern' for the sequence

Math for fun, 1, 2, 3, 4, 5, 8, ____, ____,

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#blackpenredpen #math #calculus #apcalculus

[blackpenredpen]

Yes, I just shortened the video. I will change it back to the original after a week or so. This way, people will have a chance to try it first.

[proto]

There are many possible answers:
9, 10, (11, 12, 13, 16, 17, ...)
7, 16, (9, 32, 11, 64, 13, ...)
9, 16, (17, 32, 33, 64, 65, ...)
11, 16, (21, 32, 43, 64, 85, ...) My favorite
I could list more, but I think most would recognize those as reasonable

[estuardodiaz]

You could use a cosine function: f(x) =
"Proof:"
...I used "cos(pi*x)" because it returns 1 if it is an even number and -1 if it is an odd number.
Then, we can rewrite the two equations like this: {-b+c=n}if n is odd, and {b+c=2^(n/2)} if n is even.
Solving for that you get 'b=1/2[2^(n/2)-n]' and 'c=1/2[n+2^(n/2)]

So now we have not just combined the two formulas, but also extended to all the real numbers :D
Have fun!

[kwinvdv]

You can always come up with infinitely many continuations of a sequence. For example you can fit a n-1th order polynomial on to the first n given points and get a consistent relation for all the next points. In this case the first solution I came up with was: 1, 2, 3, 4, 5, 8, 9, 16, 17, 32, 33, 64, ...
So there is not unique solution to this problem.

[HughDennin]

Solution that combines the two equations:
a(n) = (1/2) * (1 + (-1)^(x+1)) * x + (1/2) * (1 + (-1)^(x)) * 2^(n/2)

[scitwi]

Dammit, I thought of `6` and `10`. Reason: they're the missing roots to the polynomial:
x⁸ – 39·x⁷ + 633·x⁶ – 5565·x⁵ + 28854·x⁴ – 89796·x³ + 162392·x² – 154080·x + 57600
:q
Seriously though, I always sucked at these "find the missing numbers" puzzles because I could come up with just any numbers I wanted and find a justification for it that was perfectly valid to me, but perfectly wrong to the person who made the puzzle because they thought of some other, much more boring pattern :P And the thing is, I could pick any numbers at random and justify them too, because there's _always_ a polynomial that passes through these numbers and has them as its roots :P

As for your other question about joining the odd and even series into one:
Let's call the odd term `O` and the even term `E`.
We know that integer powers of `-1` alternate: even powers are `1` and odd powers are `-1`.
So if we can use `(-1)^n` and simply add it to `1` to alternate between `0` and `2` (equidistant from `1` by a positive/negative unit).
In order to get rid of the unwanted factor of `2`, just divide it by `2`. Here's the formula:
a_n = E·[1 + (-1)^n ]/2 + O·[1 + (-1)^(n+1)]/2
Or we can define two predicate functions:
isEven(n) = [ 1 + (-1)^n ] / 2
isOdd(n) = [ 1 + (-1)^(n+1) ] / 2
returning `1` when the predicate is true and `0` otherwise, and use them this way:
a_n = E·isEven(n) + O·isOdd(n)
For each `a_n`, only one of these predicates will be `1` and the other one will be `0`, so there's always only one piece of this sum, and the other one gets zeroed out.

[fourierable]

Sequence explained by <bprp> is not unique. Perfectly I can choose  1, 2, 3, 4, 5,   8, 910, 11, 12,   15, 16, 17, 18, 19,   22, 23, 24, 25, 26,   29,  …  &c.

[ribozyme]

Well, there are 238 sequences in the OEIS containing 1, 2, 3, 4, 5, 8 as a subsequence. (The first of the blanks seems to be 7 or 9 relatively often.)

[JoshuaBailey]

I saw a bunch of other formulae that looked super eloquent, so I kept trying to make mine look nicer. I think I'm happy with it now...

x + ( 2 ^ ( x / 2 ) - x ) ( cos ^2 ( π x / 2 ) )

And, more generally,

If y=f(x) when x is even, and y=g(x) when x is odd, then

g(x) + ( f(x) - g(x) ) ( cos ^ 2 ( π x / 2 ) )

and when f(x) = 2^ ( x / 2 ), g ( x ) = x, and x ϵ Z(>=0), the range this set.

Proof:

If x is even, then for some i ϵ Z(>=0), x=2 i
=> cos ^ 2 ( π 2 i / 2 ) = cos ^ 2 ( π i ) = 1
=> y = g(x) + ( f(x) - g(x) ) (1) = g(x) + f(x) - g(x) = f(x)...

If x is odd, then for some i ϵ Z(>=0), x=2 i + 1
=> cos ^ 2 ( π ( 2 i + 1 ) / 2 ) = cos ^ 2 ( ( 2 i π + π ) / 2 ) = cos ^ 2 ( i π + ( π / 2 ) ) = sin ^ 2 ( i pi) = 0
=> y = g(x) + ( f(x) - g(x) ) ( 0 ) = g(x)...

I may have made a mistake or two (it's been a long time since school. lol) but I think it worked out nice anyway. lol If anyone sees anything I could have done better, I'm happy to learn something new. =)

[DrQuatsch]

I would say the next two are 7 and 16.

[Quasarbooster]

I'll admit that I needed the hint to figure out exactly what the sequence was, but I WAS able to find a formula myself (I haven't seen the comments). It's kind of gross, but it works at least:

a_n = [(-1)^(n+1)+1]/2 * n + [(-1)^n+1]/2 * 2^(n/2)

The weird -1 stuff basically acts like a switch for integer values of n, which turns the left/right term either on or off. The rest is pretty simple.

[Schurik]

My first thought was that there are infinitely many solutions according to this view:
1 < 2 < 3 < 4 < 5 < 8 < k < k+1
We just need to say k < k+1 and k > 8. Then we can use infinitely many numbers for k and of course infinitely many for k+1.

[Fematika]

7 then 16. How I did it below:







The values at even orders are powers of two, and the other ones are odd numbers.They can't be prime because 1 is in there.

[Cuby]

A version i come up with is:
a_n =
It's based on the fact that ((-1)^n+1)/2 alternates from 0 to 1 and ((-1)^(n+1)+1)/2 alternates between 1 and 0, both starting from n=1 of course. using this technique its easy as putting the functions inside the formula above.

Simplify this term up there, it results in this:
a_n = n*(2^(n/2)/n)^(((-1)^n+1)/2) is a nice polynom, or a nice root version is: n*sqrt((2^(n/2)/n)^((-1)^n+1)), short and compact, which is my term simplified and combined together using mostly the power rule :)

A lovely general rule i come up is this formula:
a_n(E, O) = O*(E/O)^(((-1)^n+1)/2), where E is any term which should appear on even n's and O is the any term which should appear on odd n's

IMPORTANT HINT:
O can't be 0, as this would be a division by 0.
E=0 results in a series of just zeros.

Have fun trying. Had fun trying tinkering around, was alot of fun.

[admink]

How about this?
an = (n mod 2)(n) + 2^(n/2)((n + 1) mod 2)

[gdsfish]

19, 50
this is a polynomial function, it can be expressed as
x^5*1/60 - x^4*1/4 + x^3*17/12 - x^2*15/4 + x*167/30 - 2

I just made 6 equations with the six coefficients as variables
ax^5+bx^4+cx^3+dx^2+ex+f

and then I plugged in all six values,
solving it took a bit of time but now that I see the result, I am still pretty proud of it since the actual solution seems kinda cheap, especially the modulo answers which I am seeing in the comments
seems like I am the only one with that answer

[veraeron]

1) introduce an "on/off" factor F (and F') kinda like in logic gates, whose value is dependent on whether n is odd or even,
2) then pair them up with their corresponding pieces in the piece-wise defined function such that for every n, only one piece will evaluate and the other will be 0.
3) The equation will be: a sub n = F*n + F' * 2^(n/2)

1)In this case, we could just use 0 and 1 as on and off.
1.1) To make F dependent on whether n is odd or even:
F = [1 + (-1)^(n+1)]/2. Odd n: F=1. Even n: F=0 n+1 could be replaced with n-1
1.2) same goes with F':
F' = [1 + (-1)^(n)]/2. Odd n: F'=0. Even n: F'=1.

2) Equation: a sub n = [1 + (-1)^(n+1)]/2 * n + [1 + (-1)^(n)]/2 *2^(n/2)


.. This concept could be extended to more than two pieces and could also be more intricate like logic circuits using a little bit more boolean algebra.. although I think the pieces need to be periodic every time .. not sure though.

[제로-wx]

7, 16
because this can be separated to two different sequence,
for every odd order of the sequence we have just odd numbers,
for the other half of the sequence we have 2, 4, 8, it increases by 2n.

[ehtuanK]

to turn a sequence of alternating terms into one formula I use the following trick:

(1+(-1)^n)/2 is 1 if n is even and 0 if n is odd, and
(1-(-1)^n)/2 is 0 if n is even and 1 if n is odd.

so ((1+(-1)^n)* a +(1-(-1)^n)* b )/2 will equal a if n is even and b if n is odd!

After plugging in n for odd terms and 2^(n/2) for even terms we get:
((1+(-1)^n)* 2^(n/2) +(1-(-1)^n)* n )/2

Maybe you could shorten it a little bit, but having 2^(n/2) instead of 2^n makes it a bit unwieldy, so it doesn't get much better:
(2^(n/2) +(-sqrt2)^n +n -n*(-1)^n)/2

[mattwyeth]

Just add the two functions together
a(n) = n*sin((pi*n)/2)^2 +2^(n/2)*sin(pi*(n+1)/2)^2