On the invariance of the terms in A215940 (final)

(With corrections added at the end.)

    Greetings,

    Dear reader(s), your attention please...

    In the moment of writing this to all of you, I have on my table
    a scientific calculator model fx-570ES from CASIO. It allows
    to write an Algebraic expression and evaluate it in 4 different
    bases.

    I will give a practical demonstration about what is the actual
    meaning of talking about the terms of A215940 as an "Universal"
    sequence. Let me emphasize now that this is not about the
    quantity of terms or the size of the set of permutations.

    First than all, the actually important fact is the divisibility
    of the difference between two permutations by (10-1) for every
    radix or base of a positional number system where such
    permutations could be read.

    About the universality of the terms generated by
    quotients of the kind (B-A)\(10-1), for two permutations
    (A<B):

    Actually it means an invariant property of those terms
    consisting in being written always with the same numerals
    regardless the radix where they are read (of course, 1234
    cannot be written 1234 in binary, the it is clear that
    additionally we need a radix equal or greater than the
    number of objects to be permuted in order to perform the
    present experiment).

    Well, first the target: Permutations without repetitions
    in the naturals.

    1) Pick 2 different permutations without repetitions
    for the first 5 non-negative integers, or like I've
    chosen to define and call it (please see also:
    http://pastebin.com/9LNqD8ka):

    Please pick two different natural permutations for radix 5.

    Let's say 01234 and 30421.

    2) Enable the "radix or base-N" mode in your calculator,
    set it to decimal and write once the following:

    -------------------------------------------
    (30421-01234)/(10-1)


    -------------------------------------------

    When ready press "equal" (=), and see the result,
    (it usually should appears below at the right side.)

    -------------------------------------------
    (30421-01234)/(10-1)
                                           Dec
                                          3243
    -------------------------------------------

    Now without deleting the expression, change
    with just one strike to hexadecimal, and
    observe the change in the answer field:

    -------------------------------------------
    (30421-01234)/(10-1)
                                           HEX
                                      00000CAB
    -------------------------------------------

    Now without doing other things, press again
    "equal" (=). If done correctly, you will see:

    -------------------------------------------
    (30421-01234)/(10-1)
                                           HEX
                                          3243
    -------------------------------------------

    Well, don't believe me, just try it, turn
    back the calculator radix to decimal, again
    without deleting the expression, and only
    hitting one time the proper key. Look once
    the result:

    -------------------------------------------
    (30421-01234)/(10-1)
                                            Dec
                                          12867
    -------------------------------------------

    Don't be worried. Press "equals" (=)

    -------------------------------------------
    (30421-01234)/(10-1)
                                            Dec
                                           3243
    -------------------------------------------

    Arrived this point. Are you intrigued?...
    perhaps you might be so. If you now scroll
    up to the beginning text for this exercise
    here in this window or panel, you will notice
    that this result is identical to the other
    corresponding to the first time you stroked
    "equal" (=). Sure, check it.

    yes is it!

    For (A<B) any pair of distinct permutations
    without repetitions writable in your calculator,
    and the expression (B-A)/(10-1) evaluated in all
    the radices supported by your device (such that
    each radix is greater than the number of digits
    in your permutation), hitting "equal" (=) each
    time you set a different radix you will find
    that, in such base the resulting quotient is
    written always using the same "digits" or
    numerals. This and other properties are
    already documented (or the author tried to
    do this) as the sequences A215940,
    A217626 and A196020, inside the
    OEIS(TM) database (a.k.a Sloane's encyclopedia).

    Now the second part of this experiment:

    Try all the procedure described before but
    now for permutations with repetitions...

    Bingo: the "it doesn't behave in the same way",
    in fact The permutations with repetitions
    like 12343 and 12334 doesn't have this
    nice property.

*********************************************
   Essential corrections:
*********************************************

Let me advice about this... Counter-example:

Two permutations with repetitions destroy my
hypothesis: (3114 - 1314)\(10 - 1), the resulting
quotient always is written as "200".

**Then, what is happening there??: Simple,
the right explanation is, for being invariant,
the studied quotients should not contain one of
the digits that have assigned (10-1) for some
radix under consideration. Example: In decimal
the quotient (3114 - 1341)\(10 - 1) looks like
"197", then, neither octal nor decimal should
be included inside the invariance set of radices,
since (10-1)= 9 in decimal and (10-1)= 7 in octal.

This is in fact and additional restriction I didn't
considered before (to Jan 7 2012 3:01 GMT).

The invariance property is satisfied only for
those set of radices such that the considered
quotients doesn't contain the symbol for (10-1).

But it is the final result of sparse distribution
of time dedicated to analyze and explain these
matters. So we need to talk about invariance,
but defining precisely two sets: The quotients
claimed to be invariants under radix reading
interpretation, and the radices where they
are invariants.

Final comment: A215940 was always worked by me
mainly in decimal. Until tried A217626 for 11!
(base-11) permutations, where arises for
"first time" the symmetry breakdown and
the negative terms, I couldn't be able
to identify the last restriction. But
this is worthy for studying in the
further from an algebraic viewpoint
based on Diophantine series.

There it will be obvious that being:

f_{i}(N)=sum_{k=0..(N-1)}P(i,k)*10^k

A permutation, if some P(i,u) were (10-1) then:

P(i,k=u)*10^k= (10-1)*10^u= -1*10^u + 10^(u+1),

Clearly it adds a unit to another "digit"... and
subtract its corresponding power. This would be
what destroys the expected pattern of invariance.

Please also try this: (21-12)\(10-1), and after
only (21-12). The quotient always is 1, but the
difference is distinct for each radix:

Is just (10-1) in the studied radix.

This accounts for this final correction in the
explanation and interpretation given for the
present experiment.

Also historically (at least for me), upon the need
of learn how to compute generalized determinant
polynomials, the observation that in decimal 12+9=21
was the beginning of all the research.

Then for the regarded radices, each possible (10-1)
cannot be PART of one of the quotients claimed
to be invariants.

End-of-the-experiment.

Regards,

R. J. Cano, <aallggoorriitthhmmuuss@gmail.com>
Jan 7th 2013, 11:45am (VET)

Acknowledgements:
Thanks to OEIS(TM) for the patience, the support and
for being all of you good examples and source of
inspiration for get the best for our sequences.