Is it possible to get the 'canonical prime factorization', that means to get prime numbers with zero exponents too.
Example:
I have: factor(45) = 3^2 * 5
I need: factor(45) = 2^0 * 3^2 * 5
F = factor(45)
print 'F = factor(45) = ', F
print 'F[0] =', F[0]
print 'F[1] =', F[1]
Out:
F = factor(45) = 3^2 * 5
F[0] = (3, 2)
F[1] = (5, 1)
something like this:
facteurs = list((410767).factor())
print(facteurs)
[(7, 2), (83, 1), (101, 1)]
solution = facteurs
deja =[p for p,e in facteurs]
for premier in primes(facteurs[0][0]+1,stop=facteurs[-1][0]):
if premier not in deja:
solution.append((premier,0))
print(solution)
[(7, 2), (83, 1), (101, 1), (11, 0), (13, 0), (17, 0), (19, 0), (23, 0), (29, 0), (31, 0), (37, 0), (41, 0), (43, 0), (47, 0), (53, 0), (59, 0), (61, 0), (67, 0), (71, 0), (73, 0), (79, 0), (89, 0), (97, 0)]
happy new year
Asked: 2020-01-01 15:21:55 +0100
Seen: 768 times
Last updated: Jan 01 '20
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something like this:
facteurs = list((410767).factor())
print(facteurs)
[(7, 2), (83, 1), (101, 1)]
solution = facteurs
deja =[p for p,e in facteurs]
for premier in primes(facteurs[0][0]+1,stop=facteurs[-1][0]):
print(solution)
[(7, 2), (83, 1), (101, 1), (11, 0), (13, 0), (17, 0), (19, 0), (23, 0), (29, 0), (31, 0), (37, 0), (41, 0), (43, 0), (47, 0), (53, 0), (59, 0), (61, 0), (67, 0), (71, 0), (73, 0), (79, 0), (89, 0), (97, 0)]
happy new year