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

∃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?

$$\exists{x}\in\mathbb{Z}^{+} :{x}={p}_{\mathrm{1}} {p}_{\mathrm{2}} ...{p}_{{n}} \wedge\forall{p}\neq{x},\:{p}\in\mathbb{Z}^{+} \\ $$$$ \\ $$$${x}=\left({factors}\:{of}\:{p}_{\mathrm{1}} \right)\left({fact}.\:{p}_{\mathrm{2}} \right)...\left({fact}.\:{p}_{{n}} \right) \\ $$$$\mathrm{How}\:\mathrm{could}\:\mathrm{you}\:\mathrm{formally}\:\mathrm{write}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{you}\:\mathrm{can}\:\mathrm{write}\:{x}\:\mathrm{in}\:\mathrm{terms} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:{n}\:\mathrm{integers}? \\ $$

Commented by FilupSmith last updated on 27/May/16

ω(x)=number of distinct prime factors e.g. 12=2^3 ×3^1 ω(12)=2 x=Π_(t=1) ^(ω(x)) p_t ^e_t (product of prime factors) let ℧(x)=total number of prime factors ℧(x)=Σ_(t=1) ^(ω(x)) e_t ∴ combinations C(x) = (℧(x))! C(x) = (Σ_(t=1) ^(ω(x)) e_t )! ????

$$\omega\left({x}\right)=\mathrm{number}\:\mathrm{of}\:\mathrm{distinct}\:\mathrm{prime}\:\mathrm{factors} \\ $$$$\mathrm{e}.\mathrm{g}.\:\mathrm{12}=\mathrm{2}^{\mathrm{3}} ×\mathrm{3}^{\mathrm{1}} \\ $$$$\omega\left(\mathrm{12}\right)=\mathrm{2} \\ $$$${x}=\underset{{t}=\mathrm{1}} {\overset{\omega\left({x}\right)} {\prod}}{p}_{{t}} ^{{e}_{{t}} } \:\:\:\:\:\:\left({product}\:{of}\:{prime}\:{factors}\right) \\ $$$${let}\:\mho\left({x}\right)=\mathrm{total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{prime}\:\mathrm{factors} \\ $$$$\mho\left({x}\right)=\underset{{t}=\mathrm{1}} {\overset{\omega\left({x}\right)} {\sum}}{e}_{{t}} \\ $$$$\therefore\:{combinations}\:{C}\left({x}\right)\:=\:\left(\mho\left({x}\right)\right)! \\ $$$${C}\left({x}\right)\:=\:\left(\underset{{t}=\mathrm{1}} {\overset{\omega\left({x}\right)} {\sum}}{e}_{{t}} \right)! \\ $$$$???? \\ $$

Commented by FilupSmith last updated on 27/May/16

if combinations of two products x=pq x=(p_1 p_2 ...p_m )(q_1 q_2 ...q_n ) =(m+n)! =(℧(p)+℧(q))! for x=abc...n C(x)=(℧(a)+℧(b)+...+℧(n))! i think this is the case

$$\mathrm{if}\:\mathrm{combinations}\:\mathrm{of}\:\mathrm{two}\:\mathrm{products}\:{x}={pq} \\ $$$${x}=\left({p}_{\mathrm{1}} {p}_{\mathrm{2}} ...{p}_{{m}} \right)\left({q}_{\mathrm{1}} {q}_{\mathrm{2}} ...{q}_{{n}} \right) \\ $$$$=\left({m}+{n}\right)! \\ $$$$=\left(\mho\left({p}\right)+\mho\left({q}\right)\right)! \\ $$$$ \\ $$$$\mathrm{for}\:{x}={abc}...{n} \\ $$$${C}\left({x}\right)=\left(\mho\left({a}\right)+\mho\left({b}\right)+...+\mho\left({n}\right)\right)! \\ $$$$\mathrm{i}\:\mathrm{think}\:\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{case} \\ $$

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