Distinct Permutation-based Sums

[Max]

Leroy Quet wrote:

>>That's not true. a(m) may not exist. And it does not exist even for 
>>small examples.
>>Let a(1)=0, a(2)=1, a(3)=3. This sequence distinguishes all permutations 
>>of order 3.
>>But at the same time there is an equality
>>
>>a(3)*1 + a(1)*2 + a(4)*3 + a(2)*4 = a(2)*1 + a(3)*2 + a(4)*3 + a(1)*4
>>
>>that holds for ANY a(4).
>>Hence,  a(4) can NOT be defined such that a(1), a(2), a(3),a(4) 
>>distinguishes all permutations of order 4.
>>
>>Regards,
>>Max
>>
>>    
>>
>
>
>
>Uggg, you are right.
>
>So, typing without thinking...
>
>Can we establish a sequence with undistiguishable (by the definition 
>implied here) m's?
>
>We could define, for this purpose, a(m), if the m-permutation is 
>undistinguishable, arbitrarily as the lowest positive(/nonnegative) 
>integer not among a(1),a(2),...,a(m-1).
>
>  
>
Once you've got undistinguishability, it holds forever. So your 
"arbitrary" definition will work all the time, and the tail of sequence 
will be just successive integers.

>So, the undistinguished m's would form the sequence:
>
>1, 4,...
>
>Is this sequence infinite?
>(For all I know, EVERY m >= 4 is undistinguished with the a's as 
>redefined.)
>
>
>And the a-sequence then would progress as (with 0 = a(1))
>
>0, 1, 3, 2,...
>
>Ideally, the a-sequence would then form an interesting permutation of the 
>nonnegative integers. (But then we would need there to be an infinite 
>number of indistinguished m's, given how the a-sequence is now defined, I 
>believe.)
>  
>
It's simply

0, 1, 3, 2, 4, 5, 6, 7, 8, 9, 10, ...

that's not much interesting.

Regards,
Max