Powers of 2 with all even digits

COMMENTS Are there any more terms in this sequence? Evidence that the sequence may be finite, from Rick L. Shepherd , Jun 23 2002: 1) The sequence of last two digits of 2^n, A000855 of period 20, makes clear that 2^n > 4 must have n == 3, 6, 10, 11, or 19 (mod 20) for 2^n to be a member of this sequence. Otherwise, either the tens digit (in 10 cases), as seen directly, or the hundreds digit, in the 5 cases receiving a carry from the previous power's tens digit >= 5, must be odd. 2) No additional term has been found for n up to 50000. 3) Furthermore, again for each n up to 50000, examining 2^n's digits leftward from the rightmost but only until an odd digit was found, it was only once necessary to search even to the 18th digit. This occurred for 2^12106 whose last digits are ...3833483966860466862424064. Note that 2^12106 has 3645 digits. (The clear runner-up, 2^34966, a 10526-digit number, required searching only to the 15th digit. Exponents for which only the 14th digit was reach ... Read full article.