For SO(n) a calculation using LiE gives:
(using partition notation so W is [2])
and assuming n is not small
For WotimesW, [4],[3,1],[2,2],[2],[1,1],[]
(all with multiplicity one)
and for WotimesWotimesW,
1.[6] 2.[5,1] 3.[4,2] 1.[3,3] 1.[4,1,1] 2.[3,2,1] 1.[2,2,2] 3.[4] 6.[3,1] 2.[2,2] 3.[2,1,1] 6.[2] 3.[1,1] 1.[]
The same works for Sp(n) by taking conjugate partitions.
There is also a relationship with SL(n).
This is taking your question at face value. If it is understanding you're after instead then the best approach is to use crystal graphs.
The notation I have used denotes a representation by a partition. I have put m. in front to denote multiplicity is m. A partition is [a1,a2,a3,...] where aigeaj if i less than j. To convert to a highest weight vector add the appropriate number of 0s to the end.
Then take [a1−a2,a2−a3,a3−a4,...]. This gives a dominant integral weight. The fundamental weights are the partitions [1,,,,1]. If this has length k this corresponds to the k-th exterior power of the vector representation (provided 2k−1 less than n).
In particular the trivial representation is [], the vector representation V is [1], the exterior square of V is [1,1], the symmetric square is [2]+[].
For the k-th tensor power of W you will see partitions of 2k−2p for 0leplek only and it remains to determine the multiplicities (possibly 0). For SL(n) just take the partitions of 2k (with their multiplicities) and ignore the rest.
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