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There are five numbers $(a_1,a_2,a_3,a_4,a_5)$, such that they are in Arithmetic Progression.

Given that $a_1$ and $a_2$ are factorials, is there a possibility that either $a_4$ OR $a_5$ is a factorial, too?

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  • $\begingroup$ AP. == arithmetic progression? $\endgroup$ – copper.hat Apr 4 '13 at 16:25
  • $\begingroup$ Yes. Arithmetic Progression. $\endgroup$ – Inceptio Apr 4 '13 at 16:26
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There are two cases to consider: $a_1 < a_2$ and $a_1 > a_2$ (the case $a_1 = a_2$ is trivial).

If $a_1 < a_2$, then since $0 < a_2 - a_1 < a_2$, it must be that $a_2 < a_4 < 3a_2$ and $a_2 < a_5 < 4a_2$. This leaves the possibilities:

  • $a_4 = 2a_2$ which means $a_2 = 1!$, contradicting $a_1 < a_2$.
  • $a_5 = 2a_2$, similarly impossible.
  • $a_5 = 3a_2$, meaning $a_2 = 2!$, hence $a_1 = 1! = 0!$; but then the arithmetic progression does not match $a_5 = 3!$.

If $a_1 > a_2$, then $a_1 - a_2 \ge a_2$. It follows that $a_4, a_5 \le 0$.

Hence neither situation can occur, meaning that $a_4$ and $a_5$ cannot be factorials.

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The Gamma function is convex for positive arguments, so I would say no, if the A.P. is strictly increasing or decreasing.

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  • $\begingroup$ Is there any number theory approach to show there exists no progression with $a_4$ or $a_5$ being factorials? $\endgroup$ – Inceptio Apr 4 '13 at 16:29
  • $\begingroup$ @Inceptio, I am not sure right now, will post in case. $\endgroup$ – Andreas Caranti Apr 4 '13 at 16:30
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Clearly it does not occur with 1!, 2!, 3! and 4!.
Take the increasing case. If a5>a4>a2 there must be at least 2 factorials between them ie. a5 > n(n+1)a2 where n>2 $\implies$ a5>2.3.a2 => a1+4d>6(a1+d) which is impossible since d>0 for increasing. The decreasing case follows by symmetry - such an AP would be formed by reversing above case

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