Factorization is not unique.  Let p(x) be any power series with integer
coefficients beginning with 1.  Then 1/p(x) is also a power series with
integer coefficients beginning with 1.  Then any factor f(x) can be
replaced by p(x) (f(x)/p(x)).

Gene

--- "N. J. A. Sloane" <njas@research.att.com> wrote:

> 
> 
> There are lots of algorithms for factoring in Z[x].
> But what about power series with integer coefficients?
> Is there a good algorithm?  Knuth Vol 2 is silent.
> Any hints would be appreciated!
> There are two versions, depending on whether you work to
> order n, or with the full power series.
> 
> Version 1: Given
> 
>                    2       3       4        5        6        7      
>   8
> t3 := 1 + 3 x + 7 x  + 17 x  + 42 x  + 113 x  + 321 x  + 976 x  +
> 3109 x  +
> 
>            9          10           11      12
>     10258 x  + 34698 x   + 119554 x   + O(x  ),
> 
> find t1 and t2 such that t1*t2 = t3 (mod x^12).
> 
> Answer:
> > t1;
>            2      3       4       5        6        7         8      
>   9
> 1 + x + 2 x  + 5 x  + 14 x  + 42 x  + 132 x  + 429 x  + 1430 x  +
> 4862 x
> 
>               10          11           
>      + 16796 x   + 58786 x   + ...
> 
> > t2;
>              2      3      4       5       6       7       8       9 
>      10
> 1 + 2 x + 3 x  + 5 x  + 7 x  + 11 x  + 13 x  + 17 x  + 19 x  + 23 x 
> + 29 x
> 
>            11       
>      + 31 x   + ...
> 
> 
> Do any of Maple, Mma, Maxima, PARI, Magma, etc. have a command for
> factoring
> in this ring?
> 
>  
> Version 2 (harder):
> 
> Factor:  1 + 2*Sum_{m >= 1} x^(m^2) = 1 + 2x + 2x^4 + 2x^9 + 2x^16 +
> ...
> 
> Answer:  Product_{n >= 1} (1-x^2n)(1+x^(2n-1))^2
> 
> 
> Neil Sloane
> 
> 
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