The following series seems to become an arithmetic series $$a_{n+2} = a_n^2-n\cdot a_{n+1}$$ If $$a_1=3, a_2=4$$ $$(a_3=5, a_4=6, ..., a_n = n+2)$$

Can $a_n=n+2$ be derived from the original recursion equation?

And are there any other initial values that will also lead to an arithmetic series?


Of course.

If $a_n=n+2$ and $a_{n+1}=(n+1)+2$ then $$a_{n+2}=(n+2)^2-n(n +3)=n+4. $$

For the generalizing question, assume $a_n=bn+c$ for all $n$. Then $$b(n+2)+c = (bn+c)^2-n(b(n+1)+c), $$ i.e., $$bn+(2b+c)=(2bc-b-c)n+c^2$$ for all $n$, which implies $$ b=2bc-b-c,\qquad 2b+c=c^2,$$ hence $2bc=2b+c=c^2 $. One trivial solution is $b=c=0$ (i.e., $a_n=0$ for all $n$). If $c\ne 0$, we are led to $c=2b$ and $c=2$ (i.e., $a_n=n+2$for all $n$).

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    $\begingroup$ Thanks, Hagen. Would there be some way to go from $a_{n+2}=a_n^2-n\cdot a_{n+1}$ and get to $a_n=n+2$ somehow? (not assuming $a_n = bn+c$) $\endgroup$ – Kay K. Nov 24 '15 at 17:57
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    $\begingroup$ There's a mistake in the simplification of the $n^2$ terms, but the conclusion remains. $\endgroup$ – Yves Daoust Nov 24 '15 at 18:02

If the terms form an arithmetic sequence,

$$a_n=\alpha n+\beta.$$

Then by the recurrence,

$$\alpha(n+2)+\beta=(\alpha n+\beta)^2-n(\alpha(n+1)+\beta)=(\alpha^2-\alpha)n^2+(2\alpha\beta-\alpha-\beta)n+\beta^2.$$

The term in $n^2$ can only vanish when $\alpha=0$ or $\alpha=1$, which give $$\beta=-\beta n+\beta^2\implies \beta=0,$$

or $$n+2+\beta=(\beta-1)n+\beta^2\implies \beta=2.$$

There are no other possibilities.


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