## [2017 4-18] Puzzle Time

Blame Quintushalls for this.

Moderator: Kimra

### [2017 4-18] Puzzle Time

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

I dunno, something's wrong with the artwork. I can't read any of them.
GollyRojer

Posts: 61
Joined: Sun Sep 04, 2016 10:08 pm

### Re: [2017 4-18] Puzzle Time

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
rpresser

Posts: 5
Joined: Tue Oct 07, 2014 8:28 pm

### Re: [2017 4-18] Puzzle Time

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~
IRLPinkiePie

Posts: 1
Joined: Tue Apr 18, 2017 2:54 pm

### Re: [2017 4-18] Puzzle Time

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

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

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! <.< >.>
King of Ferrets

### Re: [2017 4-18] Puzzle Time

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

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
codetaku

### Re: [2017 4-18] Puzzle Time

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

### Re: [2017 4-18] Puzzle Time

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

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

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

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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