**Not a member of Pastebin yet?**

**, it unlocks many cool features!**

__Sign Up__- (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.

RAW Paste Data