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\leftline{\sevenrm AMM.Tex[let,rwf]\today}
\leftline{\copyright\sevenrm Robert W. Floyd, September 11, 1989}
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\noindent{\bf Solution of AMM Problem E3335, June/July 1989}

The recurrence $x↓{n+2} = x↓{n+1} + {x↓n\over n+2}(\ast)$ 
has the obvious solution $y↓n = n+2$, $y↓0 = 2$, $y↓1 = 3$.  We exhibit another
solution that decays more than exponentially, with the general solution and
asymptotic behavior.

Define $h↓n = 1/0! - 1/1! + 1/2!\cdots (-)↑n/(n+2)!$; $h↓n - 1/e \propto
(-)↑n/(n+3)!$

Eliminating $(-)↑n$ between
$$\eqalignno{h↓n & = h↓{n-1} + (-)↑n/(n+2)!\,\cr
\noalign{and}
h↓{n-1} & = h↓{n-2} - (-)↑n/(n+1)!,\,\cr
\noalign{we get}
(n+2)h↓n & = (n+1) h↓{n-1} + h↓{n-2},\,\cr
\noalign{from which}
z↓n & = (n+2) h↓n\,, z↓0 = 1\,, z↓1 = 1,\,\cr}$$
is another solution of $(\ast)$.

Combining these solutions,
$$\displaylines{v↓n = e z↓n - y↓n = (n+2) e \bigl (h↓n - {1\over e}\bigr ) 
\propto (-)↑n e/(n-2)!\propto (- e/n)↑n e/\sqrt{2\pi n↑5},\cr}$$
with $v↓0 = e-2, v↓1 = e-3,$ 
is the decaying solution of $(\ast)$; $v↓{12} < 10↑{-10}$.

The general solution for $x↓0 = a$, $x↓1 = b$ is 
$x↓n = \bigl ( (e-2) b - (e-3) a \bigr )(n+2)/e + (3a - 2b) v↓n/e.$

When $(e-2) b \not= (e-3) a$, 
$x↓n = {(e-2) b- (e-3) a\over e} (n+2) + O (e↑n/n↑{n+5/2}).$

When $(e-2) b = (e-3) a$, 
$x↓n = {3a - 2b\over e} v↓n\propto {(a-b) e (-e/n)↑n\over \sqrt {2\pi n↑5}}.$
\bye