Re: [math-fun] Integers with many 0s in base 2 and base 3 (fwd)

At 03:47 PM 12/21/02, ed pegg wrote:

...

Prove or disprove: There exists an epsilon>0, such that no natural number has the property that in base 2 as well as in base 3, at most (epsilon)*100% of the digits are nonzero.

Example: 513=(1000000001) in base 2 and 513=(201000) in base 3 so epsilon<1/3.

272629962 = (10000010000000000000011001010) in base 2, 272629962 = (201000000001000000) in base 3 so epsilon < 6/29

1208614932, 2453670144, 17448310278, 22083026976 all have epsilon 1/5 .

2542645806624 gives 4/21. -- Fred W. Helenius <fredh@ix.netcom.com>