x-Z-x-p-1-p-2-p-n-p-x-p-Z-x-factors-of-p-1-fact-p-2-fact-p-n-How-could-you-formally-write-the-number-of-ways-you-can-write-x-in-terms-of-the-product-of-n-integers-

Question Number 5777 by FilupSmith last updated on 27/May/16

\exists x \in Z^{+} : x = p_{1} p_{2} \dots p_{n} \land \forall p \neq x, p \in Z^{+}

x = (f a c t o r s o f p_{1}) (f a c t . p_{2}) \dots (f a c t . p_{n})

How could you formally write the

number of ways you can write x in terms

of the product of n integers ?

Commented by FilupSmith last updated on 27/May/16

ω (x) = number of distinct prime factors

e . g . 12 = 2^{3} \times 3^{1}

ω (12) = 2

x = \underset{t = 1}{\prod^{ω (x)}} p_{t}^{e_{t}} (p r o d u c t o f p r i m e f a c t o r s)

l e t ℧ (x) = total number of prime factors

℧ (x) = \underset{t = 1}{\sum^{ω (x)}} e_{t}

∴ c o m b i n a t i o n s C (x) = (℧ (x))!

C (x) = (\underset{t = 1}{\sum^{ω (x)}} e_{t})!

? ? ? ?

Commented by FilupSmith last updated on 27/May/16

if combinations of two products x = p q

x = (p_{1} p_{2} \dots p_{m}) (q_{1} q_{2} \dots q_{n})

= (m + n)!

= (℧ (p) + ℧ (q))!

for x = a b c \dots n

C (x) = (℧ (a) + ℧ (b) + \dots + ℧ (n))!

i think this is the case