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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.

February 8th, 2014, 12:06 AM   #42
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Re: Pre-Q&A

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 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.

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

February 8th, 2014, 12:17 AM   #44
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Re: Pre-Q&A

Quote:
 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 ...

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.

 February 8th, 2014, 12:39 AM #46 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 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.
 February 8th, 2014, 12:43 AM #47 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory 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.
 February 8th, 2014, 04:53 AM #48 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: Pre-Q&A Have managed a slight improvement. $f(x) \= \ (x^2 - 3x + 3)(x^6 + 3x^5 + 6x^4 + 9x^3 + 9x^2 + 8x + 3 ) \ =$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x^8 + 8x^3 + 6x^2 + 15x + 9$ f(2) = 383 So we are in {0 ... 15}
 February 8th, 2014, 04:58 AM #49 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Pre-Q&A Superb. {0, ..., 15} we are in now then
 February 8th, 2014, 05:10 AM #50 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 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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