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I would appreciate if somebody could help me with the following problem:

Q: How to find $a_n=?$

$$(n+1)a_{n+2}-na_{n+1}-a_{n}=0,a_1=1, a_2=0$$

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deleted by user21820, rschwieb, RRL Apr 2 at 14:47

closed as off-topic by user223391, JonMark Perry, Lord Shark the Unknown, Joel Reyes Noche, Wouter Feb 21 '18 at 5:15

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Hint: Let us multiply both sides by $n!$

$(n+1)!a_{n+2}=n(n!a_{n+1})+n((n-1)!a_{n})$

We now let $(n-1)!a_n=b_n$

$b_{n+2}=n(b_{n+1}+b_n)$

Now, it is a recurrence form similar to derangement.

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  • $\begingroup$ Could you give a reference to what you call a derangement ? $\endgroup$ – Jean Marie May 7 '16 at 12:33
  • $\begingroup$ en.wikipedia.org/wiki/Derangement I believe this should do it. $\endgroup$ – Sinpoint May 7 '16 at 12:36
  • $\begingroup$ A quick computation with Mathematica confirms that $a_n=1/2, 2/3, 9/4, 44/5, 265/6, 1854/7, 14833/8,...$ etc... whose numerators are members of the OEIS sequence A000166 oeis.org of derangements. $\endgroup$ – Jean Marie May 7 '16 at 12:54

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