if you know enough about the logic of your program to hardcode a==1 or a==7 or a==15, then you know enough to hardcode set([1,7,15]) outside the loop.

Creating the set takes ~6 times longer than doing one hardcoded comparison, but further comparisons are quite a bit faster. I think the choice of what to use depends on how many times you’ll execute the statement. But there should never be a case you create the set repeatedly.

In [1]: a=16

In [2]: s=set([1,7,15])

In [3]: timeit a==1 or a==7 or a==16
10000000 loops, best of 3: 126 ns per loop

In [4]: timeit a in s
10000000 loops, best of 3: 107 ns per loop

It took me a minute to figure out what you were doing with the list, but after reading a comment or two I realized that I had to do this for one number.
x=10
if x in (10,):
print ‘hello’

I suppose there’s more than one way to skin a cat.