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Powers of 2 with all even digits
OFFSET 1,1 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.
(The clear runner-up, 2^34966, a 10526-digit number, required searching only to the 15th digit. *) 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 *)
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