$n$th Mathematics Blog

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Product of Odd Factorials: Do They Exist?

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Case 1 ($n=1$): $$x_1!=1!\cdot3!=6=3!$$ Case 2 ($n=2$): $$x_2!=1!\cdot3!\cdot 5!=(1)\cdot(1\cdot 2\cdot 3)\cdot (1\cdot 2\cdot 3\cdot 4\cdot 5)=1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6=6!$$ Case 3 ($n=3$): $$x_3!=1!\cdot3!\cdot 5!\cdot 7!=(1)\cdot(1\cdot 2\cdot 3)\cdot (1\cdot 2\cdot 3\cdot 4\cdot 5) \cdot (1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7)$$ $$=1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8\cdot 9 \cdot 10=10!$$ Case 4 ($n=4$): $$x_4!=1!\cdot3!\cdot 5!\cdot 7!\cdot 9!$$ It is sufficient to say that: $$x_4!>11!\implies 11\mid x_4!$$ However: $$11 \mid 1!\cdot3!\cdot 5!\cdot 7!\cdot 9!$$ Case 5 ($n=5$): $$x_5!=1!\cdot3!\cdot 5!\cdot 7!\cdot 9!\cdot 11!$$ It is sufficient to say that: $$x_5!>13!\implies 13\mid x_5!$$ However: $$13 \mid 1!\cdot 3!\cdot 5!\cdot 7!\cdot 9!\cdot 11!$$ Case $n$ ($n \ge 6$): $$1!\cdot3!\cdot 5! \cdots (2n+1)!=x_{\ge 6}!$$ It is sufficient to say that: $$x_{\ge 6}!>(4n+2)!$$

Bertrand’s Postulate (should be a theorem) states that $\forall n>1, n< p< 2n$ for some prime, $p$. Generalizing this to our argument, we have the following: $$2n+1< p < 4n+2$$ $$\implies (2n+1)!< p!<(4n+2)!$$ $$\implies (2n+1)!< p!<(4n+2)!< x_{\ge 6}!$$ $$\implies p \mid x!$$ However, since $p!>(2n+1)!$, there exists no $p$ such that $p \mid x!$. Therefore, $p \mid x!$.

Proposition does not hold past $n=1,2,3$. So, no they do not exist.

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