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

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

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

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

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