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Question Number 96749 by student work last updated on 04/Jun/20

how we can calclate triple factorial?

$$\mathrm{how}\:\mathrm{we}\:\mathrm{can}\:\mathrm{calclate}\:\mathrm{triple}\:\mathrm{factorial}? \\ $$

Answered by Rio Michael last updated on 04/Jun/20

The tripple factorial is defined as n!!! = n(n−3)(n−6)...3, n(n−3)(n−6)...4∗1 or n(n−3)(n−6)...5∗2 depending on the numbers congruency so n!^((a)) should be defined as the product of numbers congruent to n(mod a) as examples : 8 (mod 3) ≡ 5 (mod 3) ≡ 2 (mod 3) n!^((a)) =n(n−a)(n−2a)...(m +2a)(m + a)m ,where m is the smallest possible number allowed by the factorial.

$$\mathrm{The}\:\mathrm{tripple}\:\mathrm{factorial}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{as}\: \\ $$$$\:{n}!!!\:=\:{n}\left({n}−\mathrm{3}\right)\left({n}−\mathrm{6}\right)...\mathrm{3},\:\:{n}\left({n}−\mathrm{3}\right)\left({n}−\mathrm{6}\right)...\mathrm{4}\ast\mathrm{1} \\ $$$$\mathrm{or}\:{n}\left({n}−\mathrm{3}\right)\left({n}−\mathrm{6}\right)...\mathrm{5}\ast\mathrm{2}\:\:\mathrm{depending}\:\mathrm{on}\:\mathrm{the}\:\mathrm{numbers}\:\mathrm{congruency} \\ $$$$\mathrm{so}\:{n}!^{\left({a}\right)} \:\mathrm{should}\:\mathrm{be}\:\mathrm{defined}\:\mathrm{as}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{numbers}\:\mathrm{congruent}\:\mathrm{to}\:{n}\left(\mathrm{mod}\:{a}\right) \\ $$$$\mathrm{as}\:\mathrm{examples}\::\: \\ $$$$\mathrm{8}\:\left(\mathrm{mod}\:\mathrm{3}\right)\:\equiv\:\mathrm{5}\:\left(\mathrm{mod}\:\mathrm{3}\right)\:\equiv\:\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{3}\right) \\ $$$${n}!^{\left({a}\right)} \:={n}\left({n}−{a}\right)\left({n}−\mathrm{2}{a}\right)...\left({m}\:+\mathrm{2}{a}\right)\left({m}\:+\:{a}\right){m}\:,\mathrm{where}\:{m}\:\mathrm{is}\: \\ $$$$\mathrm{the}\:\mathrm{smallest}\:\mathrm{possible}\:\mathrm{number}\:\mathrm{allowed}\:\mathrm{by}\:\mathrm{the}\:\mathrm{factorial}. \\ $$

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