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February 8th, 2014, 12:05 AM   #41
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Re: Pre-Q&A

Quote:
Originally Posted by agentredlum
I was thinking the same thing about putting it in the notes
See the reduction on the interval I gave. I essentially derived it from your example.
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February 8th, 2014, 12:06 AM   #42
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Re: Pre-Q&A

Quote:
Originally Posted by mathbalarka
Using agent's factorization, by reduction of the constant term, I get

x^7 + 5x^2 + 12x + 15 = (x^5 + 3x^4 + 6x^3 + 9x^2 + 9x + 5)(x^3 - 3x + 3)

so record is {0, ..., 15} Infinitesimal improvements. Dang.

EDIT I realized that agent took the binomial I have gave before and multiplied it by x and added a 9 to get the second part going. Nice!
I got that too but unfortunately f(2) = 11*17

No worries though...it's still ineresting as a factorisation.

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February 8th, 2014, 12:10 AM   #43
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Re: Pre-Q&A

Quote:
Originally Posted by agentredlum
I got that too but unfortunately f(2) = 11*17
Right. It's hard to keep track of the conditions
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February 8th, 2014, 12:17 AM   #44
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Re: Pre-Q&A

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Originally Posted by mathbalarka
I realized that agent took the binomial I have gave before and multiplied it by x and added a 9 to get the second part going. Nice!
Its nice you found the shortcut , i didn't do that , actually went the long way by trial and error , so i can't take credit.

But ...

If what you say is true we may be sniffing at a structure similar to the J commutative monoid.

Heh he , imagine that ...

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February 8th, 2014, 12:26 AM   #45
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Re: Pre-Q&A

Quote:
Originally Posted by agentredlum
Its nice you found the shortcut , i didn't do that , actually went the long way by trial and error , so i can't take credit.
I am wondering whether similar can be done with my {0, ..., 16} example to reduce the tuple further. One thing is clear : The bigger factor, the smaller the coefficients.
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February 8th, 2014, 12:39 AM   #46
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Re: Pre-Q&A

If you can do that thn you will probably solve the OP entirely. I'll try it too since it looks promising.

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February 8th, 2014, 12:43 AM   #47
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Re: Pre-Q&A

I have tried it but no dice. A negative term pops up. It's interesting that gigantic constructions are possible :

(x^3 - 3x + 3)(x^8 + 4x^7 + 10x^6 + 19x^5 + 27x^4 + 25x^3 - 5x^2 - 89x - 251) = x^10 + x^9 + x^8 + x^7 + x^5 + x^4 + x^3 + x^2 + 486x - 753

Whereas we have trouble with the small ones.
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February 8th, 2014, 04:53 AM   #48
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Re: Pre-Q&A

Have managed a slight improvement.






f(2) = 383

So we are in {0 ... 15}

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February 8th, 2014, 04:58 AM   #49
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Re: Pre-Q&A

Superb. {0, ..., 15} we are in now then
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February 8th, 2014, 05:10 AM   #50
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Re: Pre-Q&A

It looks like something magical happens when we hit consecutive coefficients of 9. You know , I found smaller coefficients using your 'generator polynomial' , (x^2 - 3x + 3) and working on the sextic , got it down to {0 ... 12} but unfortunately f(2) was not prime.

Next up would be to try a 7th degree ...

agentredlum predicts most coefficients of the 7th degree poly ,

1 3 6 9 9 a b c

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