[2017 4-18] Puzzle Time

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Moderator: Kimra

[2017 4-18] Puzzle Time

Postby GollyRojer » Tue Apr 18, 2017 3:19 pm

http://www.smbc-comics.com/comic/puzzle-time

I dunno, something's wrong with the artwork. I can't read any of them.
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Re: [2017 4-18] Puzzle Time

Postby rpresser » Tue Apr 18, 2017 3:32 pm

y = 0.9303569123 x^7 - 22.23193879 x^6 + 207.829112 x^5 - 961.3816793 x^4 + 2289.732664 x^3 - 2625.385296 x^2 + 1111.506783 x + 0.00001611345942

Graph on Wolfram Alpha
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Re: [2017 4-18] Puzzle Time

Postby IRLPinkiePie » Tue Apr 18, 2017 4:53 pm

i got it as (521/560)x^7 - (2587/90)x^6 + (43291/120)x^5 - (170393/72)x^4 + (2085781/240)x^3 - (6369907/360)x^2 + (638962/35)x - 7219, but to each their own, i guess~
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Re: [2017 4-18] Puzzle Time

Postby Woom » Tue Apr 18, 2017 5:23 pm

My version is essentially the same, but rendered as:
( 4689 * n^6 - 112049 * n^5 + 1047459 * n^4 - 4845365 * n^3 + 11540256 * n^2 - 13231946 * n + 5601996 ) * n / 5040

(The next term is 8666.)
Woom
 

Re: [2017 4-18] Puzzle Time

Postby Cryft » Tue Apr 18, 2017 5:54 pm

https://oeis.org/search?q=0%2C1%2C4%2C-13%2C-133%2C52%2C53%2C-155&sort=&language=english&go=Search

If he was just randomly thinking "I wonder what sequence hasn't been used before..." then he really nailed it.
Cryft
 

Re: [2017 4-18] Puzzle Time

Postby King of Ferrets » Tue Apr 18, 2017 9:52 pm

Psssh, that's easy.

f(0) = 1
f(x) = 0 when x != 0

f(x-1)*1+f(x-2)*4+f(x-3)*-13+f(x-4)*-133+f(x-5)*52+f(x-6)*53+f(x-7)*-155

See? Simple! <.< >.>
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Re: [2017 4-18] Puzzle Time

Postby Guest » Tue Apr 18, 2017 10:20 pm

I'm personally fond of the form
f(n) = ₙC₁ + 2 ₙC₂ − 22 ₙC₃ − 61 ₙC₄ + 552 ₙC₅ − 1940 ₙC₆ + 4689 ₙC₇.
Guest
 

Re: [2017 4-18] Puzzle Time

Postby codetaku » Wed Apr 19, 2017 1:18 am

Woom wrote:My version is essentially the same, but rendered as:
( 4689 * n^6 - 112049 * n^5 + 1047459 * n^4 - 4845365 * n^3 + 11540256 * n^2 - 13231946 * n + 5601996 ) * n / 5040

(The next term is 8666.)


I just used matlab to get an estimate of the coefficients via polyfit (obviously I had to specify degree 7 since that's the "simplest" polynomial that fits these parameters, though there are infinitely many polynomials of greater degree that also fit this sequence).

Anyway, I concur that the next term is 8666 in the case that we assume the simplest possible pattern fit to the sequence. But let's be honest, there's no author intention because I doubt zach knows how to fit a polynomial to a set of points :P
codetaku
 

Re: [2017 4-18] Puzzle Time

Postby ThePaulanator » Wed Apr 19, 2017 7:34 am

Someone should definitely submit it on oeis as the weinersmith numbers.
ThePaulanator
 

Re: [2017 4-18] Puzzle Time

Postby cmena2702 » Wed Apr 19, 2017 8:34 am

The obvious solution is the sequence defined as follows:

The sequence an is defined by:
{a1 = 0
a2 = 1
a3 = 4
a4 = -13
a5 = -133
a6 = 52
a7 = 53
a8 = -155} for an from N.

Yes, I am a dickhead. My area is pure mathematics; answers that are technically true but completely meaningless are what I do best.
cmena2702
 

Re: [2017 4-18] Puzzle Time

Postby codetaku » Wed Apr 19, 2017 1:01 pm

cmena2702 wrote:The obvious solution is the sequence defined as follows:

The sequence an is defined by:
{a1 = 0
a2 = 1
a3 = 4
a4 = -13
a5 = -133
a6 = 52
a7 = 53
a8 = -155} for an from N.

Yes, I am a dickhead. My area is pure mathematics; answers that are technically true but completely meaningless are what I do best.


I am willing to believe that your area is pure mathematics because anyone *sane* would be 0-indexing this sequence ;P but regardless, you seem to have missed the comma+ellipsis in the original sequence. This directly implies that the sequence has length greater than 8. Many even use it to imply that the sequence is infinite (perhaps repeating, but still infinite), but at the very least, not even a theoretical mathematician would put an ellipsis after the end of a finite sequence.
codetaku
 

Re: [2017 4-18] Puzzle Time

Postby DrHammer » Wed Apr 19, 2017 3:43 pm

I'll take my 14 points please.

{n∈Z|(∃x∈N)[n=H(x-1)+3H(x-2)-17H(x-3)-120H(x-4)+185H(x-5)+H(x-6)-208H(x-7)]}
Where Z are the integers, N is the set of natural numbers, and H is the Heaviside function.
DrHammer
 

Re: [2017 4-18] Puzzle Time

Postby codetaku » Wed Apr 19, 2017 6:37 pm

DrHammer wrote:I'll take my 14 points please.

{n∈Z|(∃x∈N)[n=H(x-1)+3H(x-2)-17H(x-3)-120H(x-4)+185H(x-5)+H(x-6)-208H(x-7)]}
Where Z are the integers, N is the set of natural numbers, and H is the Heaviside function.


Hahah, okay, I bow to you, that was a perfect answer.

And all future elements are -155. So elegant! You won't even introduce nonintegers!
codetaku
 


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