A new aproach to C.Collatz. 3n+1...
 

For all Collatz series, starting with any odd n, always meets the equation:

(2^M)- P = n*(3^STEPS)

With M,P functions of n.

JM M

Does this tell us anything more than "the number of steps is a function of n"?

Conjetura de Collatz-Ulam. (Lothar Collatz. Stanislaw Ulam)

Definición. Se define la función productoPotencia, , como la suma de productos de D, en binario, por una secuencia de potencias de tres. Se ignoran los bit cero.

proPOT(D) = SUM( (3^I)*(2^J) ).

Ejemplo. D=151 =128+16+4+2+1.
proPOT(151) =SUMA[(1,3,9,27,81)*(128,16,4,2,1)] =347.

Collatz-Ulam es la t-serie con la ec.de paso: 3*e+1 = (2^g)*e' , donde N=0, eFinal=1, y eInicial el valor inicial.
Ec. equivalente a la función de Syracuse. Syracuse(e)=e'.

Ejemplo con eInicial =7
e: 7,11,17,13,5,1. g: 1,1,2,3,4. M=11. D= 1+2+4+16+128 = 151.

Ec. de intervalo: (2^11) = proPOT(151) + (7)*(3^5)

Siendo:
proPOT(151)= SUMA[(1,3,9,27,81)*(128,16,4,2,1)] =347.
#pasos =bits(D) =5

Teorema
Para la serie de Collatz se obtiene:

(2^Collatz_M(eInicial)) =proPOT(D) + (eInicial)*(3^bits(D))

y la conjetura equivale a afirmar que para todo n, existe un d tal que:

2^Collatz_M(n) -proPOT(D) = n*3bits(D)

Algunas soluciones
M D n 3^bits(D)
(2^ 5) = proPOT( 3) + 3 * (3^ 2)
(2^11) = proPOT( 151)+ 7 * (3^ 5)
(2^13) = proPOT( 605)+ 9 * (3^ 6)
----------------------------------------------------------------------


Share Knowledge.

JM M

[QUOTE=JM Montolio A;481040]
Share Knowledge.
[/QUOTE]
[COLOR=LemonChiffon].[/COLOR]

Mire, Sr. Sergev Batalov. Replique a mis post ó no los conteste.
Pero yo no entro aquí para las tonterias. De nadie.

JM Montolio

Why do [B]you[/B] post nonsense then?
You should not take nonsense from anyone. Including yourself.

And what, are you denying other people the right to not take nonsense from anyone?

The water drip observation was aimed at all of your odd dozen last threads, which are peppered all over the forum all of a sudden and all of which are
[B]"New"
"Useful"[/B]
etc.
Even the forum software recognizes you as a spammer.

Please learn some modesty and learn to listen to others.

[QUOTE=VBCurtis;481035]If you'd stop calling your trivial observations "useful", you (and the forum) would be much better off.[/QUOTE]

If one post is NOT NEW, please tell about it,
and can be deleted. No problem.

If one post is NOT USEFUL, same thing.

No problem on corrections.

Delete, move.

But with respect.

JM M

[QUOTE=JM Montolio A;481096]If one post is NOT NEW, please tell about it,
and can be deleted.
...

Delete, move.
[/QUOTE]
What you are describing is censorship.
... Also can you kindly explain how someone can "delete with respect"?

What I am describing is free speech.

P.S. Btw, in case you are not understanding it - everyone with [COLOR="Red"]nicknames[/COLOR] in [COLOR="Red"]red[/COLOR] is a moderator. You are talking to a moderator. A moderator who had enough dealing with spam.

Dr. Batalov, le agradezco su paciente comprensión.
 

Reciba usted mis mejores grifos.


JMMA

Nothing new.
If you apply the Condensed Collatz function [TEX]\frac{3e+1}{2^g}[/TEX] starting from 7 until you reach 1, you get
[TEX]11=\frac{3\cdot7+1}{2^1}[/TEX]
[TEX]17=\frac{3\cdot(\frac{3\cdot7+1}{2^1})+1}{2^1}=\frac{3^2}{2^2}\cdot7+\frac{3^1}{2^2}+\frac{3^0}{2^1}[/TEX]

...

[TEX]1=\frac{3^5}{2^{11}}\cdot 7+\frac{3^4}{2^{11}}+\frac{3^3}{2^{10}}+\frac{3^2}{2^9}+\frac{3^1}{2^7}+\frac{3^0}{2^4}[/TEX]

or rearranged:

[TEX]7=\frac{2^{11}}{3^5} - (\frac{2^7}{3^5}+\frac{2^4}{3^4}+\frac{2^2}{3^3}+\frac{2^1}{3^2}+\frac{2^0}{3^1})[/TEX]

multiply by [TEX]3^5[/TEX] and you get

[TEX]7\cdot 3^5=2^{11} - (2^7\cdot 3^0+2^4\cdot 3^1+2^2\cdot 3^2+2^1\cdot 3^3+2^0\cdot 3^4)[/TEX]
or
[TEX]7\cdot 3^5=2^{11} - (128\cdot 1+16\cdot 3+4\cdot 9+2\cdot 27+1\cdot 81)[/TEX]

all you do is rearrange the whole thing.