Eric Angelini wrote:
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:
<snip>
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
In a way, this is also consistent and appropriate, as 2 is -1 mod 3.
(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
Why 2 is not marked just as 12 (i.e. +1*3 + -1*1 ?) And then 3 = 10, 4 = 11 as you have given, but 5 = +1*9 + -1*3 + -1*1 = 122 6 = +1*9 + -1*3 + 0*1 = 120 7 = +1*9 + -1*3 + 1*1 = 121 8 = +1*9 + 0*3 + -1*1 = 102 9 = +1*9 + 0*3 + 0*1 = 100 I am sorry, but the terms 1,12,10,11,122,120,121,102,100 do not match anything in the table.
The "ternary notation" above can certainly be bettered as the "ternary artefacts" 20, 1220, represent nothing.
No. They represent negative integers: -1 = -1*1 = 2 -2 = -1*3 + 1*1 = 21 -3 = -1*3+0*1 = 20 -4 = -1*3 + -1*1 = 22 -5 = -1*9 + 1*3 + 1*1 = 211 -6 = -1*9 + 1*3 + 0*1 = 210 -7 = -1*9 + 1*3 + -1*1 = 212 -8 = -1*9 + 0*3 + 1*1 = 201 -9 = -1*9 + 0*3 + 0*3 = 200 We see that to convert an integer to its negative in this notation, we have just to swap ones and twos 1 <-> 2, and keep zeros intact. I am sorry, but the terms 2,21,20,22,211,210,212,201,200 do not match anything in the table. Now, of course any such ternary string (either 0, or beginning with a non-zero digit, i.e. either 1 or 2) represents non-negative integers in the usual ternary notation: http://www.research.att.com/projects/OEIS?Anum=A007089 And because I'm looking for bijections everywhere, even simple ones, one expects two to be lurking here. We just need to "fold" the integers to natural numbers with f: Z -> N as given by f(z) = 2z if z>0 else 2|z|+1, and vice versa, with its inverse* *g(z) = z/2 if z even else (1-z)/2. So we get those two sequences (for positive and negative integers) interleaved: I am sorry, but the terms 0,1,2,12,21,10,20,11,22,122,211,120,210,121,212,102,201,100,200 do not match anything in the table. Reinterpreting them in ordinary ternary notation, we get: I am sorry, but the terms 0,1,2,5,7,3,6,4,8,17,22,15,21,20,16,23,11,19,9,18, do not match anything in the table. And actually, this seems to be an involution (self-inverse permutation), so we get just one permutation, not two. Why is this? (Or is it?) I have to think about it, but first I will finish my coffee to raise my IQ temporarily from that of the greater apes. Salut, Regards, Groetjes, Terveisin, Antti
Best, Ã.
(seq. 0,1,121,10,11,1211,1210,12121,1201,100,... is not in the OEIS)