 # On the invariance of the terms in A215940 (final)

Jan 7th, 2013
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1. (With corrections added at the end.)
2.
3.     Greetings,
4.
6.
7.     In the moment of writing this to all of you, I have on my table
8.     a scientific calculator model fx-570ES from CASIO. It allows
9.     to write an Algebraic expression and evaluate it in 4 different
10.     bases.
11.
12.     I will give a practical demonstration about what is the actual
13.     meaning of talking about the terms of A215940 as an "Universal"
14.     sequence. Let me emphasize now that this is not about the
15.     quantity of terms or the size of the set of permutations.
16.
17.     First than all, the actually important fact is the divisibility
18.     of the difference between two permutations by (10-1) for every
19.     radix or base of a positional number system where such
21.
22.     About the universality of the terms generated by
23.     quotients of the kind (B-A)\(10-1), for two permutations
24.     (A<B):
25.
26.     Actually it means an invariant property of those terms
27.     consisting in being written always with the same numerals
29.     cannot be written 1234 in binary, the it is clear that
31.     number of objects to be permuted in order to perform the
32.     present experiment).
33.
34.     Well, first the target: Permutations without repetitions
35.     in the naturals.
36.
37.     1) Pick 2 different permutations without repetitions
38.     for the first 5 non-negative integers, or like I've
40.     http://pastebin.com/9LNqD8ka):
41.
43.
44.     Let's say 01234 and 30421.
45.
47.     set it to decimal and write once the following:
48.
49.     -------------------------------------------
50.     (30421-01234)/(10-1)
51.
52.
53.     -------------------------------------------
54.
55.     When ready press "equal" (=), and see the result,
56.     (it usually should appears below at the right side.)
57.
58.     -------------------------------------------
59.     (30421-01234)/(10-1)
60.                                            Dec
61.                                           3243
62.     -------------------------------------------
63.
64.     Now without deleting the expression, change
65.     with just one strike to hexadecimal, and
66.     observe the change in the answer field:
67.
68.     -------------------------------------------
69.     (30421-01234)/(10-1)
70.                                            HEX
71.                                       00000CAB
72.     -------------------------------------------
73.
74.     Now without doing other things, press again
75.     "equal" (=). If done correctly, you will see:
76.
77.     -------------------------------------------
78.     (30421-01234)/(10-1)
79.                                            HEX
80.                                           3243
81.     -------------------------------------------
82.
83.     Well, don't believe me, just try it, turn
84.     back the calculator radix to decimal, again
85.     without deleting the expression, and only
86.     hitting one time the proper key. Look once
87.     the result:
88.
89.     -------------------------------------------
90.     (30421-01234)/(10-1)
91.                                             Dec
92.                                           12867
93.     -------------------------------------------
94.
95.     Don't be worried. Press "equals" (=)
96.
97.     -------------------------------------------
98.     (30421-01234)/(10-1)
99.                                             Dec
100.                                            3243
101.     -------------------------------------------
102.
103.     Arrived this point. Are you intrigued?...
104.     perhaps you might be so. If you now scroll
105.     up to the beginning text for this exercise
106.     here in this window or panel, you will notice
107.     that this result is identical to the other
108.     corresponding to the first time you stroked
109.     "equal" (=). Sure, check it.
110.
111.     yes is it!
112.
113.     For (A<B) any pair of distinct permutations
114.     without repetitions writable in your calculator,
115.     and the expression (B-A)/(10-1) evaluated in all
117.     each radix is greater than the number of digits
118.     in your permutation), hitting "equal" (=) each
119.     time you set a different radix you will find
120.     that, in such base the resulting quotient is
121.     written always using the same "digits" or
122.     numerals. This and other properties are
123.     already documented (or the author tried to
124.     do this) as the sequences A215940,
125.     A217626 and A196020, inside the
126.     OEIS(TM) database (a.k.a Sloane's encyclopedia).
127.
128.     Now the second part of this experiment:
129.
130.     Try all the procedure described before but
131.     now for permutations with repetitions...
132.
133.     Bingo: the "it doesn't behave in the same way",
134.     in fact The permutations with repetitions
135.     like 12343 and 12334 doesn't have this
136.     nice property.
137.
138. *********************************************
139.    Essential corrections:
140. *********************************************
141.
143.
144. Two permutations with repetitions destroy my
145. hypothesis: (3114 - 1314)\(10 - 1), the resulting
146. quotient always is written as "200".
147.
148. **Then, what is happening there??: Simple,
149. the right explanation is, for being invariant,
150. the studied quotients should not contain one of
151. the digits that have assigned (10-1) for some
152. radix under consideration. Example: In decimal
153. the quotient (3114 - 1341)\(10 - 1) looks like
154. "197", then, neither octal nor decimal should
155. be included inside the invariance set of radices,
156. since (10-1)= 9 in decimal and (10-1)= 7 in octal.
157.
158. This is in fact and additional restriction I didn't
159. considered before (to Jan 7 2012 3:01 GMT).
160.
161. The invariance property is satisfied only for
162. those set of radices such that the considered
163. quotients doesn't contain the symbol for (10-1).
164.
165. But it is the final result of sparse distribution
166. of time dedicated to analyze and explain these
167. matters. So we need to talk about invariance,
168. but defining precisely two sets: The quotients
170. interpretation, and the radices where they
171. are invariants.
172.
173. Final comment: A215940 was always worked by me
174. mainly in decimal. Until tried A217626 for 11!
175. (base-11) permutations, where arises for
176. "first time" the symmetry breakdown and
177. the negative terms, I couldn't be able
178. to identify the last restriction. But
179. this is worthy for studying in the
180. further from an algebraic viewpoint
181. based on Diophantine series.
182.
183. There it will be obvious that being:
184.
185. f_{i}(N)=sum_{k=0..(N-1)}P(i,k)*10^k
186.
187. A permutation, if some P(i,u) were (10-1) then:
188.
189. P(i,k=u)*10^k= (10-1)*10^u= -1*10^u + 10^(u+1),
190.
191. Clearly it adds a unit to another "digit"... and
192. subtract its corresponding power. This would be
193. what destroys the expected pattern of invariance.
194.
195. Please also try this: (21-12)\(10-1), and after
196. only (21-12). The quotient always is 1, but the
197. difference is distinct for each radix:
198.
199. Is just (10-1) in the studied radix.
200.
201. This accounts for this final correction in the
202. explanation and interpretation given for the
203. present experiment.
204.
205. Also historically (at least for me), upon the need
206. of learn how to compute generalized determinant
207. polynomials, the observation that in decimal 12+9=21
208. was the beginning of all the research.
209.
210. Then for the regarded radices, each possible (10-1)
211. cannot be PART of one of the quotients claimed
212. to be invariants.
213.
214. End-of-the-experiment.
215.
216. Regards,
217.
218. R. J. Cano, <aallggoorriitthhmmuuss@gmail.com>
219. Jan 7th 2013, 11:45am (VET)
220.
221. Acknowledgements:
222. Thanks to OEIS(TM) for the patience, the support and
223. for being all of you good examples and source of