Hello SeqFan and Math-Fun, Consider this (hope this is not old-hat): 3^0 3^1 3^2 3^3 3^4 3^5 3^6 3^7 ... = 1 3 9 27 81 243 729 2187 ... This seq. is known to be very efficient if you want to weigh integer weights with a two-tray balance, leaving no "holes" behind; use weights of 1,3,9,27... units to measure all "natural" quantities, from 1 to infinity: 1 = 1 2 = 3-1 3 = 3 4 = 3+1 5 = 9-(3+1) 6 = 9-3 7 = 9-(3-1) 8 = 9-1 9 = 9 10 = 9+1 11 = 9+(3-1) 12 = 9+3 13 = 9+(3+1) 14 = 27-[9+(3+1)] etc. I had the idea, yesterday night, to represent all natural numbers in the same way -- all I had to do was to use three symbols : 0 for a power of 3 I don't need 1 for a power of 3 I need 2 as a grahic symbol meaning "minus all the rest" ... and use a regular addition rule (which will be explained below in a few seconds). Let's make first a parallel with the binary no- tation : 2^0 2^1 2^2 2^3 2^4 2^5 2^6 2^7 ... = 1 2 4 8 16 32 64 128 ... We use two symbols : 0 for a power of 2 we don't need 1 for a power of 2 we need ... and an easy addition rule. We start from the left, putting 0's and 1's where we want to, and then reverse the sequence of 0's and 1's. To represent 54 (my age) we look for the closest power < or = to 54 and mark it with a "1": 2^0 2^1 2^2 2^3 2^4 2^5 2^6 2^7 ... = 1 2 4 8 16 32 64 128 ... --> 1 54 - 32 = 22; we mark with another "1" the closest power < or = to 22: 2^0 2^1 2^2 2^3 2^4 2^5 2^6 2^7 ... = 1 2 4 8 16 32 64 128 ... --> 1 1 54 - (32+16) = 6, we mark accordingly 2^2: 2^0 2^1 2^2 2^3 2^4 2^5 2^6 2^7 ... = 1 2 4 8 16 32 64 128 ... --> 1 1 1 54 - (32+16+4) = 2, we mark 2^1: 2^0 2^1 2^2 2^3 2^4 2^5 2^6 2^7 ... = 1 2 4 8 16 32 64 128 ... --> 1 1 1 1 To complete the binary representation of 54, we put 0's under all unnecessary powers: 2^0 2^1 2^2 2^3 2^4 2^5 2^6 2^7 ... = 1 2 4 8 16 32 64 128 ... --> 0 1 1 0 1 1 ... last line is reversed to produce 110110, which is indeed the binary expression of 54. We can represent all natural integers with this method, of course: 2^0 2^1 2^2 2^3 2^4 2^5 2^6 2^7 ... = 1 2 4 8 16 32 64 128 ... 1 = 1 => 1 2 = 0 1 (to reverse)=> 10 3 = 1 1 => 11 4 = 0 0 1 => 100 5 = 1 0 1 => 101 6 = 0 1 1 => 110 7 = 1 1 1 => 111 8 = 0 0 0 1 => 1000 9 = 1 0 0 1 => 1001 etc. Let's do the same with the powers of 3; we will mark with a "1" the powers we need, with a "0" the ones we will discard and (novelty!) with a "2" the quantity we will substract (I will use the star symbol (*) to indicate hereunder an operation in base 10): 3^0 3^1 3^2 3^3 3^4 3^5 3^6 3^7 ... = 1 3 9 27 81 243 729 2187 ... 1 = 1 2 = *3-1* = 121 3 = 01 => 10 4 = *3+1* = 11 5 = *9-4* = 1211 6 = *9-3* = 1210 7 = *9-2* = 12121 8 = *9-1* = 1201 9 = 001 => 100 10 = *9+1* = 101 11 = *9+2* = 1121 12 = *9+3* = 110 13 = *9+4* = 111 14 = *27-13* = 12111 15 = *27-12* = 12110 16 = *27-11* = 121121 17 = *27-10* = 12101 18 = *27-9* = 12100 19 = *27-8* = 121201 20 = *27-7* = 1212121 21 = *27-6* = 121210 22 = *27-5* = 121211 23 = *27-4* = 12011 24 = *27-3* = 12010 25 = *27-2* = 120121 26 = *27-1* = 12001 27 = 0001 => 1000 28 = *27+1* = 1001 29 = *27+2* = 10121 30 = *27+3* = 10010 etc. Is this of interest? Is the introduction of a third symbol "economi- cally" efficient? (I think we will use less symbols to represent all first 1000 natural integers with this "ternary notation" than with the usual binary system -- the price might be to high, though)... Speaking of "notation efficiency" (how to re- present in the less "symbol-consuming" way all the natural numbers) -- is there a definitive answer? The "ternary notation" above can certainly be bettered as the "ternary artefacts" 20, 1220, 001 or 002 represent nothing. Best, É. (seq. 0,1,121,10,11,1211,1210,12121,1201,100,... is not in the OEIS)