The 4^n Polynomial conjecture

[cino hilliard]

>From: "Marc LeBrun" <mlb at fxpt.com>
>Reply-To: <mlb at fxpt.com>
>To: "'Edwin Clark'" <eclark at math.usf.edu>,        "'cino hilliard'" 
><hillcino368 at hotmail.com>
>CC: "'seqfan'" <seqfan at ext.jussieu.fr>, <ltownley2003 at yahoo.com>,        
><hilliard.t at comcast.net>
>Subject: RE: The 4^n  Polynomial conjecture
>Date: Mon, 17 Nov 2003 09:38:41 -0800
>
> >= Edwin Clark
> >>= Cino Hilliard
>
> >>Polynomial numbers of the form 4^n + 4^(n-1) + ... + n mod 2 + 1
>
> > The formula is not clear. I think you mean:
> >   (4^n + 4^(n-1) + ... + 4 + 1) + (n mod 2)
> > Is this correct?

Yes. This is an interpretation of what I intended. I was being brief. If n 
is even 1 is the constant term
else it is 2. To me 4^n = 4^(n-1) ... implies expanding the polynomial to 
the point of 4^(n-n) = 1
The addition n mod 2 is MY arbitrary choice to create a sequence of a 
particular character..


>
>I agree that this is a much better way to write it--that mod in the
>penultimate term is confusing (as is the adjective "polynomial").
The n mod 2  + 1 is MY formula to toggle between 1 and 2 depending on 
whether n is even or odd.
I thoughrt it rather clever to avoid a if clause. Eg.,
Instead of my  Polynomial numbers of the form  4^n + 4^(n-1) + ... + n mod 2 
+ 1
try
Polynomial numbers of the form  4^n + 4^(n-1) + ... + r  where r = 1 if n is 
even or r=2 if n is odd.
6 to .5 DZ to the other.

To me polynomial means "many names" or many terms. I see no problem using 
poloynomial in this
context.



>
>
>And the repunit part of course sums to the expression (4^(n+1)-1)/3
>(ie (b^(n+1)-1)/(b-1) for other values of 4)

Does the repunit create my sequence?

>
> > Unless I misunderstand something here's a proof of the conjecture:
> > If n is odd then a(n) = 4^n + ... + 4^1 + 2 so is even and not prime.
> > If n is even then a(n) = 4^n + ... + 4^1 + 1, that is, is a repunit
> > to base 4 and according to Chris Caldwell's repunits page at
> > http://primes.utm.edu/glossary/page.php?sort=GeneralizedRepunit
> > the only one is 4^1 + 1. But since 1 is not even that's not a problem.

Thanks. I did a google search for 4^n, 4^(n-1) etc without meaningful hits. 
Must have missed it.

>
>This isn't exactly a proof, because that page just claims to cover
>numbers of less than 100 digits, while commenting "(You might want to
I took 4^n + 4^(n-1) + . . . 4^1  + n mod 2 +1 to n=10000 and found no 
pseudoprimes
(16:53) gp > f(n) = s=0;for(x=1,n,s+=4^x);length(Str(s))
(16:57) gp > f(100)
%3 = 61
(16:57) gp > f(1000)
%4 = 603
(16:57) gp > f(10000)
%5 = 6021

That is in excess of 6000 digit numbers.


>explain why the lists for b=4, 9, 16 and 25 are so short.)" which
>suggests that there may be a theorem for square b.
I will check it out.
>
> > It look like another situation of probability -
> > the larger the numbers get the scarcer the primes.
>
>You're shooting logarithmically scarcer fish in an exponentially bigger
>barrel.  It's not surprising that we hit very few primes.  On the other
>hand, it's also not a proof that you won't accidentally nail one
>somewhere on the way to infinity.  But the compositeness doesn't seem
>remarkable.

Does to me. Hey, I am just having fun! Recall Fermat was shooting when he 
assumed 2^32+1
was prime. It was beleived for some 95 years to be true  by some of the 
greatest minds
until Euler factored it in his head to be  641*6700417. Imagine that. any 
8th grader could have done
it  by trial division of the 119 primes <= 641. Say 3 division (checking) = 
5 minutes. 119*5 =  595
minutes =~ 10 hours! spread it. a class of 20 could have done int in 1/2 
hour!
This seems remarkable to me because for 95 years so many thought it not to 
the point of not
questioning or doing it.

These are the primes involved in the project. If any one is a grade school 
teacher this would
be a nice project to sharpen the paper and pencil mind or (ARRGH) 
calculator. Watch out though,
some one in the class may use Pari,Maple,etc Third graders?


%18 = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 
67, 7
1, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 
151, 15
7, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 
239, 24
1, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 
337, 34
7, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 
433, 43
9, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 
541, 54
7, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 
641]



Have fun.
cino

BTW, When you reply back to the list, just send to. do not cc the names. I 
keep getting
double posts.
>

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