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3N Integration Algorithm #575

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I think I have figured it out! It appears to be a modified version of the 3S* algorithm from Ketcheson (2010). Removing the error calculation and adding an extra term to S2 when I=5. This seems to produce the equations from Gottlieb (2009) exactly:

$m = 5$ (number of stages)
$S_{3}=U_{i,j,k}^{n}$
$S_{2}=0$
$S_{1} = U_{i,j,k}^{n}$
For $I=2,~I\leq m+1,~I= I+1$ {
$S_{2} = S_{2}+\delta_{I-1}S_{1}$
If $I==5$ {
$S_{2} = S_{2} + \beta_{extra} \Delta t \cdot \vec{\nabla}[F_{i,j,k}(S_{1})]$
} End if
$S_{1} = \gamma_{I,1}S_{1}+\gamma_{I,2}S_{2}+\gamma_{I,3}S_{3}+\beta_{I,I-1}\Delta t \vec{\nabla}\cdot[F_{i,j,k}(S_{1})]$
} End for
$U_{i,j,k}^{n+1}=S_{1}$

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@BraydenAlbery
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@felker
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Answer selected by BraydenAlbery
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