Question on cyclotomic polynomials

[Edwin Clark]

I found lots more examples with 3 odd primes. Also I looked for
counter-examples to the other question about multiples a little
bit--unsuccessfully. So the second question is still open, but
if I had to bet I'd say that if Phi_n(x) has a coefficient different
from 0,1,-1 then so does Phi_{kn}(x) for all k. According to
formula (30) at 

  http://mathworld.wolfram.com/CyclotomicPolynomial.html

Phi_(n) has the same nonzero coefficients as Phi_{pn} when p is
a prime divisor of n. Which means that the "conjecture" holds
if k is any product of prime divisors of n. There have been
several papers (see refs in the above MathWorld entry) on
the magnitude of the coefficients of Phi_n(x). There may
be more there about your question. 

  

On Fri, 19 May 2006 franktaw at netscape.net wrote:

> Sorry, the word "odd" got left out; that should have been "at least 3 distinct odd prime factors".  However, 231 is still an exception.
>  
> This still leaves the second question: can you ever have Phi_n having a coefficient that is 2 or larger (in absolute value), while Phi_{k*n} does not?  A quick look at A117318 and A117223 suggests that it is not, but this is not definitive.
>  
> Franklin T. Adams-Watters
>  
> -----Original Message-----
> From: Edwin Clark eclark at math.usf.edu
> 
> On Fri, 19 May 2006 franktaw at netscape.net wrote:
> 
> > Somebody on this list must know the answer to this question.
> >  
> > It is well known that the cyclomatic polynomial Phi_n of order n can have a 
> coefficient with absolute value greater than 1 only if n has at least 3 distinct 
> prime factors.  The question is, is this if and only if?
> >  
> 
> No, if n = 2*3*5 or 3*7*11 then Phi_n(x) has coefficients 0,1,-1 only. 
> 
> Maple calculates Phi_n(x) for such values of n pretty fast.
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---------------------------------------------------------
  W. Edwin Clark, Math Dept, University of South Florida
           http://www.math.usf.edu/~eclark/
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