Thanks John,
That sorted me out, sometimes I just can't get things worked out in my head,
then get a sense of "instant enlightenment", which your comments did for me.
I am ashamed to say I was using the wrong prime factors function, then
changing the mult to mul all started to make sense.
Thanks ag
2009/1/29 col speed :
[...]
> What I expected "mult" to do was (somehow)to work out what the powers of
> the prime factors would be. Another reason I didn't think it was "mul" is
> the part that says " prime_factors_mult(n)" as the prime_factors function
> is just "prime_factors(n)" - without th
hat *should *work out the number of coprimes of 144
*Please don't bother too much about this. I've included it for your
information as syou have replied, but I think I'll leave it until I
understand a bit more - I'm biting off more than I can chew.*
Message: 6
Date: Wed,
"col speed" wrote
I got the following function while googling:
def totient(n):
from operator import mult
if n == 1: return 1
return reduce(mult, [(p-1) * p**(m-1) for p,m in
prime_factors_mult(n)])
I already have the "prime_factors" function. The problem is that I
cannot
find "mu
That's what I thought , but I tried it to no avail. Plus the syntax is
wrong.
Thanks anyway
Colin
2009/1/28 John Fouhy
> 2009/1/28 col speed :
> > Hello there,
> > I got the following function while googling:
> >
> > def totient(n):
> > """calculate Euler's totient function.
> >
> > If
2009/1/28 col speed :
> Hello there,
> I got the following function while googling:
>
> def totient(n):
> """calculate Euler's totient function.
>
> If [[p_0,m_0], [p_1,m_1], ... ] is a prime factorization of 'n',
> then the totient function phi(n) is given by:
>
> (p_0 - 1)*p_
Hello there,
I got the following function while googling:
def totient(n):
"""calculate Euler's totient function.
If [[p_0,m_0], [p_1,m_1], ... ] is a prime factorization of 'n',
then the totient function phi(n) is given by:
(p_0 - 1)*p_0**(m_0-1) * (p_1 - 1)*p_1**(m_1-1) * .
