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2n-2 days

If there are n holes,then

sequence of inspection should be,
Part 1: 2,3,4...,(n-1);
followed by
Part 2: (n-1),(n-2),...2.

if vampire started in an even numbered hole, then it will be found in part 1.
otherwise
if vampire started in odd numbered
hole , then it will be found in part 2.

Proof: The end holes 1 and n are not checked because, before being caught in these end holes, vampire will be found in hole 2 or hole n-1. in both parts we are traversing from one end to other and parity of our movement is different in part 1 and part 2. Hence hermit will be caught in whichever part our parity matches with that of vampire.

Thanks

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