I'm slightly surprised FindLinearRecurrence
does not ahndle this. But it has been a while since I have seen anything in that code base. It might be done using FindSequenceFunction
as in an example in the ref guide for DifferenceRoot
under Properties & relations.
First get the DifferenceRoot
solution.
ff =
FindSequenceFunction[{2732, 2728, 1368, 680, 344, 168, 88, 40, 24, 8,
8, 0, 4}, n]
Out[692]=
DifferenceRoot[
Function[{\[FormalY], \[FormalN]}, {-\[FormalY][\[FormalN]] + \
\[FormalY][1 + \[FormalN]] + 2 \[FormalY][2 + \[FormalN]] ==
0, \[FormalY][1] == 2732, \[FormalY][2] == 2728, \[FormalY][3] ==
1368}]][n]
Now recast as an explicit recurrence.
First[Head[ff]][a, n]
(* Out[694]= {-a[n] + a[1 + n] + 2 a[2 + n] == 0, a[1] == 2732,
a[2] == 2728, a[3] == 1368} *)