Consider the simpler problem of no consecutive heads and also no consecutive tails.  After the first flip, the next 19 flips are fully determined.  So there are only 2 possible patterns.

Suppose you do 18 flips under these new rules.  There are only 2 ways.  There are 9 Heads and 9 Tails.  If you select any 2 of those Heads and double them, you have accounted for a way of doing 20 flips under the actual rules.  So if we, somewhat arbitrarily select 18, this accounts for 2*Combin(9,2) ways of achieving our goal of 20 flips.

For an odd number 2k+1, there might be k Heads or k+1 Heads.

We can figure this for numbers from 20 down until there are not enough Heads to double to construct 20:

20+0 2*C(10,0)

19+1 C(10,1) + C(9,1)

18+2 2*C(9,2)

17+3 C(9,3) + C(8,3)

16+4 2*C(8,4)

15+5 C(8,5) + C(7,5)

14+6 2*C(7,6)

13+7 C(7,7) 

  for 13 there is only one way, 1st coin must have been Heads

The sum of all these combinations should be the answer:   465

465 is in agreement with Charlie's result which is always a good sign.