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# User talk:R. J. Cano

## April 2017

Probably acquired it (ignoring if it is a bad one) the habit of massively mailing my thoughts... However perhaps they have their worth as realizations. Example: [[Apparently having more to do with A056944][1]]

## February 3rd 2017; Thinking on pencils and the Mersenne numbers

Experiment: "A teacher instructs the students of a class for picking an arbitrarily long pencil initially including (or not) its eraser, and cutting (it or its pieces) n-1 times into two smaller pieces, sharpening both sides (if possible) of each resulting piece. The eraser could be alternatively kept or removed if desired".

Problem: Find an unique expression which sum all the pencil tips obtained with each step, after n-1 steps of the procedure previously described.

Solution: Given "e" and "t" two bits corresponding to the answers for: Is the eraser initially pr"e"sent?, and Must the eraser be kep"t"? (1=yes, 0=no) then the asked sum "s" is: s = 2*A000225(n)-e*(1+t*(n-1));

The proof about the sum is striaghtforward, as follows: If we sum all the first powers of 2 starting from exponent 0,

${\displaystyle 2^{0}+2^{1}+2^{2}+2^{3}+...2^{n}=1+2+4+8+...=\sum _{k=0}^{n}2^{k}=2^{\left(n+1\right)}-1}$

And we subtract 2 from both sides:

${\displaystyle \sum _{k=0}^{n}2^{k}-2=2^{\left(n+1\right)}-2-1=2\cdot \left[2^{n}-1\right]-1=2\cdot A000225\left(n\right)-1}$

We are done and by observation can deduce the other cases/possibilities, combining them in an unique expression.

## A humble Christmas gift... and: Happy new 2017!

[[Experimental sourcecode][2]] On those numbers F(n,b), being F hypotetically another sequence focused in base b (here b=10), such that F(n) and rev(F(n)) are different (which implies that F(n) is NOT a palindrome when expressed in base b) and F(n) is the least integer such that abs(F(n)-rev(F(n))) is a (n+1)st power;

As a particular result coming from the study of such family of definitions, we now have A278930;

## From May until Nov ( Good Bye 2016!!! ): REBOOT

The Steinhaus-Johnson-Trotter algorithm probably is the more "auto-similar" among the 3 known algorithms with such [[properties in the first differences for their respective outputs][3]] (N. Pandit, S. Zaks, Steinhaus-Johnson-Trotter);

A full & precise (interpretative) description of these observations will be reported here, soon. However temporarily: [[This is just what was found, possibly NOT yet properly explained. This is an ongoing study which might be far from being concluded...][4]]

[[ Click this to get the preceding (mostly unmodified) PARI-GP research script on these matters.][5]]

Nice distraction: Messing up with the so called "Knight's tour problem" ...All a true challenge (IMHO). Particularly, looking for a way to solve it faster purely by picking terms from an integer sequence. After all, why don't think it more like a physicist by taking advantage from its "natural" symmetries... ... ... Conceptually for now there's nothing meaningful to contribute from my part. Just this (PARI-GP sourcecode implementation) [[experimental backtracking idea! (very slow) ][6]]

## Apr 24 2016

Partial and nicely confirmed that each one of the 3 conjectures at A217626 also holds (the first with a minor and justified modification) for analogue sequence definitions (to A215940 and A217626) by replacing the Narayana Pandita's algorithm for generating permutations in increasing sequence with the Schmuel Zaks's algorithm described by a comment at A055881;

The following featured link points to a PARI-GP source code file where they are explained as comments some experiments that were performed countless times with different setup and variation of parameters, in order to find counter-examples for the any of the three conjectures at A217626. Once routines got finally correct: NO EXCEPTIONS WERE FOUND; The statements are correct, yet after replacing the permutations generating algorithm, in the way described previously and having into account (of course) the proper scaling factors that are revealed with such modification. More details inside [ this file ][7]].

## Mar 05 2016

(EXPERIMENTAL) On very elementary number theory: Let run the snowball down along the way.

Today I woke up quite intrigued. Let us see, I was asked recently to prove by either mathematical induction or modulo congruences that:

30 | n^5 - n

For every natural n;

Well, by factorization we find out that it contains as one of its factors: "The product of three consecutive naturals", which is divisible by 2, 3, and of course 6.

Driven by such observation, we could try to prove that it is also 0 (mod 10) which would complete the proof.

That's true by "Fermat's Little Theorem". MFH 14:45, 5 March 2016 (UTC)

Informal and additionally we could say that it must be true "10 | n^5 - n" since for positive integers n, and k>=0, n^(5^k) and n both end with the same digit, so "10 | n^5 - n" which would imply both conditions "2 | n^5 - n" and "5 | n^5 - n";

Curiously it is involved here a simple Diophantine equation: 2*a+1 == 4*b+1, with infinitely many solutions of the form a=2*b, and b=1 the particular case for 4*b+1 == 5;

Onwards, after shedding some light with the aid of a program (for instance a little [[GP script][8]]), we are tempted to finally state the following conjecture:

  For any pair of positive integers "x" and "y"
the number "x^y" is "x (mod 10)" if 
and only if
 "y"
has either the form:

"(4*p+1)^q" with odd "q",

Or the form:

"(2*p+1)^q" with even "q",

for positive integers p and q;



I am sure it is a true theorem, but I truly ignore about how to prove it rigorously.

No, at least the "only if" is not true since e.g. for x=1 this holds for any y. MFH 14:48, 5 March 2016 (UTC)
Danke Maximilian. R. J. Cano 15:10, 5 March 2016 (UTC)

## Jan 24 2016

Yet inclusively try this: With a little bit of Algebra, the preceding general formula (Jan 23) allow us to figure out this:

${\displaystyle \phi =\sum _{k=0}^{\infty }\phi ^{-2k}}$

Certainly a curious property of the Golden number.

## Jan 23 2016

${\displaystyle N=\left[\!2\!\cdot \!\sum _{k=0}^{\infty }\left(\!{\frac {N\!+\!1\!-\!2{\sqrt {N}}}{N\!-\!1}}\right)^{k}-1\right]^{2}}$

[[Main motivation and "plausible side-effect purpose" here.][9]]

Also a remarkably particular case (N=5, when properly rewritten):

${\displaystyle \phi ={\frac {1+{\sqrt {5}}}{2}}=\sum _{k=0}^{\infty }\left({\frac {3-{\sqrt {5}}}{2}}\right)^{k}=\sum _{k=0}^{\infty }\left(1+\varphi \right)^{k}}$

## Jan 01 2016

Happy new year everyone!,


... Thanks to Mr. [[Jon Awbrey][10]] for mentioning this interesting [[fact][11]]

Reading it inspired no few thoughts and also driven me to some "curious" observations (imho) and I would like to share with you dear visitor. Please see them @ my Talk page.

## Oct 07 2015

All a mess with modular arithmetics the problem about scheduling a book for reading it. [[This script acceptably helps in solving such kind of problems.][12]]

## Oct 06 2015

My first practical classroom work solution, as a formal student of Mathematics at the university. English approx. statement: "A reader who picks a book containing 290 pages is planning to read it 4 pages per day and 21 extra pages on sundays. By starting a sunday, how many? days, weeks are necessary to read it all, and which day-of-the-week is when finishes??".

(Answer: "41 days", this is: "5 weeks and 6 days", finishing on a Friday)

Just yesterday!,

Final and happily solved the challenge [[About: Problème 125 Envelopper un parallélépipède][13]]

[[Plain text file containing PARI-GP source code and the solution, also featuring extra related content.][14]]

## Sept 17 2015

Some nice observations, [[About reactangles composed with identical squares.][15]]

## Sept 24 2015

[[About: Problème 125 Envelopper un parallélépipède][16]]

Reboot: Temporarily working only [[the 2D version][17]] of the same problem.

## Sept 17 2015

More is told about all this work related with A008277 and so on, please see my discussion page. Thanks!

I had to be right there: A computer program carefully designed and properly implemented cannot lie!. The sort of "theorem" I was looking to find out behind some arithmetic formulas, "is real", it does exist. The Stirling numbers of the second kind as factors for a series expansion of interger powers over the integers,,, are necessarily a consequence of such Bigger theorem.

[[002) The Pascal draft which was translated in order to get #001][18]]

Partial solution...

[[001) Of the problem described by link #000, first clean hit.][19]]

I decided to share it, finally releasing this idea, so any other individual or group also interested may explore it. Please let me know if you find out something more on this...

[[000) Alternative to the multinomial theorem][20]]

## Previously

If you enter http://www.oeis.org and use my name enclosed between quotes or "R. J. Cano" as the search key, you'll obtain in the reply a list of contributions where I was involved.

At purpose of collaboration and submission, I was recently trying this since June 18 2015:

[[About semi-primes and the modern calendar.][21]]

But I have not submitted yet anything on this subject, at least not in the form of new sequences.

(That stuff was got just by curiosity.

...It looks to me very improbable to find oneself performing such kind of calculations precisely a Jun 18th of a non leap year).

...Beyond that, there is no too much to say about me.

I used to be a Physics student. Now I'm withdrawn from there, in order to study formally Mathematics.

My new interest and passion is the Riemman Hypothesis and the Complex Analysis.

## Below lies my: Old Talk page (Welcome)

Before, 2015 (Although it actually comes to born as "untitled" idea since Dec 27 2010)

${\displaystyle \delta _{1}\cdot \delta _{2}=\delta _{12}+{\cancel {\delta }}_{12}}$

This expression has a beautiful logic when it is interpreted in terms of summations, beyond the inner product of vectors and also beyond the Einstein's summation convention.

Please see the output of [[this program][22]] when it is compiled with its "Stirling2ndKind" flag set to zero.

## 2016 and the smallest Pythagorean triple

Pattern to apply: ($^$+$)^$ (where \$ are arbitrary integers)

(2^0+1)^6, gives a perfect square!

(6^1+0)^2, another perfect square, by construction.

And... both sum another perfect square.

Now an algorithmic pattern:.

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

"The Stilts rule"

Statement: Pick an arithmetic expression and remove from it (if possible) every pair of identical negative integers and the common absolute value.

Sum all the absolute values that were also removed, and multiply such sum by two.

Finally apply it as exponent for a power affecting the result of what remains from the original expression.

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

Now consider the expression: 2+0+1+6

It can be re-written as: 2+1-1+2-1+6

Now it can be applied "The Stilts rule", which gives:

(2+2+6)^2 = 10^2 = 100

Precisely the third perfect square that was missing previously.

Therefore, simple arithmetical expressions built using only the digits in 2016, and the execution of a simple algorithm like what is called here "The Stilts rule" allows us to write the 3 squares corresponding to the Pythagorean triple (6,8,10).

This is, the smallest Pythagorean triple (3,4,5) scaled by the smallest possible integer factor 2.

Curiously 2016 is also a term belonging A171476, where 6 and 8 appears as coefficients in the title / defining formula.

## The binary representation for 2016

It is: "11111100000"

Six 1s, and five zeros?. Well: Curiously you can verify that 6 and 2016 share the pattern "having such a binary representation containing almost the same quantity of zeros and units, differing only by an "1", where all the 1s are grouped at the left side", which is also nicely explained by a comment from Arthur Jasinski at A006516.

And... such A-number!: If we replace the "A" with its "Latin Alphabet" reverse index (this is enumerate from "Z" to "A"), and sum all the digits distinct than 6, and also remove repeated digits, what left is precisely a 6, and 8, the coefficients for the defining formula at A171476 which is another interpretation of the same sequence.

## On the prime 89189

Isn't it curious to have such pattern?.

If you were allowed to doodle around it arbitrarily, look: (89)1(89) is a palindrome in both base 90 and base 99 where depending how you read it (to left/right), either (89) or (98) would be the greatest "digits" or "letters";

More than that, such combination, the unit, and "base" minus 1, with both cases, as their digits...

Curiously, our previous choice "(89)1(89)", this is, to glue 8 and 9 from both sides reading them as a digit, let us call it C=89,

"C1C" in the base 90 is the same than 721079 in decimal, also a prime.

Yet think this, "C1C" with our previous choice (C=89), in the base 99 is also a prime!!!... yes: identical to 872477 in decimal.

And both ways of reading "C1C" differs by another prime plus 1;

If we call "s" to such prime, then it this true: floor(sqrt(s)) is also a prime......

However, the OEIS preferred base is the decimal. so we leave this comment here just like an arithmetical curiosity.

A nice and last extra: 89*1*89 is 7921, and then, now that we are conjecturing and flying in the kingdom of "interesting" patterns beyond the arithmetical (Number Theory) significance, we could notice as a subtlety that talking about palindromes has a lot to do with permutations: 7921 is not a prime, but 1297, a permutation of its digits, is prime.... and Ramanujan's Taxi cab 1729 is yet another permutation.

Yet more fun?, maybe: 98*1*98 is not a prime, but reversing its digits becomes a curious semi-prime: 13*313, now: Imagine you set C=13, then instead of the mentioned semi-prime we could propose to have: C+"3C" now asking for those pairs of positive integers (C,y) such that when evaluated (and alternatively* divided once by 2 if possible without getting a prime), the expression C+"3C" in the base "y" is a semi-prime (of course in the sake of soundness we should restrict it to C <= y-1).

[ Btw: Would it be (C,y=10) enough nice and interesting for becoming part of the OEIS?... well honestly not as an independent seq since it would lack of a missing term in order to have at least 4 terms, indeed: For y=10, and C in {2,4,8} the expression: C+"3C" evaluated in the base "y" gives a semi-prime, which means to have semi-primes of the form 3*y+2*C for C <= y-1 and y>1 ].

...It was a pleasure to comment this. Happy new year 2016 to everyone, a moment for which we must await 45 years more to be in 2061 an almost* a semi-prime year (due 2061 is 3*687 and 687 is the semi-prime 3*229)... predicted perihelion of Halley's Comet is 28 July 2061, and Ramanujan's Taxi cab almost match: "1729" could be read:

"1-7-29", "1" for "2061 is 1 mod 10", the "7" for July, and "29" for Friday July 29th, a good day for buying the newspaper and looking in the science section.

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

P.S.: Warming up once more, here is an stub of pari-gp code for C+"3C" eval. in the base "y" filtering only semi-primes:

/* Begin */
list3cplusc(a=10,b=10)={my(s);for(y=a,b,for(C=0,y-1,s=3*y+2*C;
/* if(!(s%2),s/=2); */
if(issquarefree(s)&&(#factor(s)[,1]==2),print1(", "y":"C,"("s")"))))}
/* End */


Where it could be noticeable the fact it gives different results if the corresponding part "divided once by 2 if possible" is executed (by leaving it uncommented).