COMMENTS
The corresponding exponents are: 1, 2, 3, 6, 11.
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 reached were only 590, 3490, 8426, 16223, 27771, 48966 and 49519 - representing each congruence above.)
No additional terms up to 2^100000. - Harvey P. Dale, Dec 25 2012
No additional terms up to 2^(10^11). See C program in links. It was only necessary to check the rightmost 40 digits of each power. But for going towards 10^12, 40 digits won't suffice. - Lukas Huwald, Mar 20 2025
No additional terms up to 2^(10^13). See second C program in links. The champions in this range are 2^133477987019 with 46 even last digits and 2^9780164740006 with 45 even last digits. - Fredrik Johansson, Mar 21 2025
No additional terms up to 2^(10^17). See Go program in links which uses a strong sieve to accelerate the computation substantially. - Ted E Dunning, Mar 31 2025
MATHEMATICA
(*returns true if none of digits of n are odd, false o.w.*) f[n_] := Module[{ a, l, r, i}, a = IntegerDigits[n]; l = Length[a]; r = True; For[i = 1, i <= l, i++, If[Mod[a[[i]], 2] == 1, r = False; Break[ ]]]; r] (*main routine*) Do[p = 2^i; If[f[p], Print[p]], {i, 1, 10^4}]
Select[2^Range[0, 100], Union[Take[DigitCount[#], {1, -1, 2}]]=={0}&] (* Harvey P. Dale, Dec 25 2012 *)
Select[2^Range[0, 100], AllTrue[IntegerDigits[#], EvenQ]&] (* The program uses the AllTrue function from Mathematica version 10 *) (* Harvey P. Dale, Sep 18 2016 *)
PROG
(PARI) f(n)=n=vecsort(eval(Vec(Str(n)))%2, , 8); #v==1&&v[1]==0
m=Mod(1, 10^19); for(n=1, 1e5, m*=2; if(f(lift(m))&&f(2^n), print1(2^n", "))) \\ Charles R Greathouse IV, Apr 09 2012