10 for N=1 to 3000
20 if N=fnSod(2^N) then print N,2^N
30 next
999 end
1000 fnSod(X)
1010 local S$,Tot,I
1015 Tot=0
1020 S$=cutspc(str(X))
1030 for I=1 to len(S$)
1040 Tot=Tot+val(mid(S$,I,1))
1050 next
1060 return(Tot)
finds only
n 2^n
5 32
70 1180591620717411303424
Why no more?
There doesn't seem to be a specific reason ruling out further cases. The number of digits in the power should vary linearly with the power and should be about n multiplied by the common log of 2. The "typical" s.o.d. should also then be proportional to n. And the maximum possible sod would be 9*n*log(2), and as small as needed.
Perhaps it's just the great range and standard deviation in the sod that make it more and more unlikely to be a "hit" as the powers get larger.

Posted by Charlie
on 20160527 15:12:52 