Ah,
I think I see what you're doing. Thanks for the explanation.
Cheers,
Tim.
On Dec 3, 2008, at 6:04 AM, Simon King wrote:
Dear Tim,
perhaps the following is a better explanation.
In the above situation, "for X in F" yields a list of pairs (x-1,2),
(x
+1,2) etc. In particular, X is n
Does this help?
sage: n=factorial(30)
sage: [Factorization([pe]) for pe in n.factor()]
[2^26, 3^14, 5^7, 7^4, 11^2, 13^2, 17, 19, 23, 29]
sage: x = polygen(GF(3))
sage: f = cyclotomic_polynomial(120)(x)
sage: [Factorization([pe]) for pe in f.factor()]
[(x^4 + x^2 + x + 1)^2,
(x^4 + x^2 + 2*x
Dear Tim,
perhaps the following is a better explanation.
In the above situation, "for X in F" yields a list of pairs (x-1,2), (x
+1,2) etc. In particular, X is not a polynomial. It is a pair, formed
by a polynomial and a number.
Hence, I am *not* applying a function called "Factorization" to so
Dear Tim,
On Dec 3, 10:57 am, Tim Lahey <[EMAIL PROTECTED]> wrote:
-- snip --
> Yes, and it's what someone else told me earlier in this thread.
Where? I only found your own suggestion [(i^j).factor() for i,j in f].
But this is likely not a good idea, because
* i^j is computed (may take time),
On Dec 3, 2008, at 4:50 AM, Simon King wrote:
Dear Tim,
On Dec 3, 7:15 am, Tim Lahey <[EMAIL PROTECTED]> wrote:
No, because I want instead of something like
[(x-2,2),(x-3,3)]
I'd like
[(x-2)^2,(x-3)^3]
You may do this:
Start with a factorization of something:
sage: f=factor(16200)
Dear Tim,
On Dec 3, 7:15 am, Tim Lahey <[EMAIL PROTECTED]> wrote:
> No, because I want instead of something like
> [(x-2,2),(x-3,3)]
>
> I'd like
> [(x-2)^2,(x-3)^3]
You may do this:
Start with a factorization of something:
sage: f=factor(16200)
sage: f
2^3 * 3^4 * 5^2
"for X in f"
On Dec 3, 2008, at 1:03 AM, Craig Citro wrote:
Thanks. That works, but Sage automatically expands things so you
need to do,
[(i^j).factor() for i,j in f]
To ensure that each polynomial term is shown factorized.
Do you want a list of pairs of the form (p,e) for each term in the
factorizati
> Thanks. That works, but Sage automatically expands things so you
> need to do,
> [(i^j).factor() for i,j in f]
>
> To ensure that each polynomial term is shown factorized.
>
Do you want a list of pairs of the form (p,e) for each term in the
factorization of f? (That's actually how the factoriza
On Dec 3, 2008, at 12:18 AM, Jason Grout wrote:
An easier way:
sage: f=factor(16200)
sage: f
2^3 * 3^4 * 5^2
sage: [i^j for i,j in f]
[8, 81, 25]
Thanks. That works, but Sage automatically expands things so you
need to do,
[(i^j).factor() for i,j in f]
To ensure that each polynomial term
Tim Lahey wrote:
> Hi,
>
> Does the Factorization class not have a
> routine to return just a list of the factors?
> Basically, something like the value function
> but instead of multiplying them, puts
> each term into a list?
>
> I ask because the output of a squarefree_decomposition
> is a fac
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