A new aproach to C.Collatz. 3n+1...

JM Montolio A · 2018-02-27, 12:28

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

CRGreathouse · 2018-02-27, 14:28

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

JM Montolio A · 2018-02-27, 15:33

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

Batalov · 2018-02-27, 15:44

Quote:

Originally Posted by JM Montolio A

Share Knowledge.

.

JM Montolio A · 2018-02-27, 17:47

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

JM Montolio

Batalov · 2018-02-27, 18:11

Why do you 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?

Batalov · 2018-02-27, 20:00

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
"New"
"Useful"
etc.
Even the forum software recognizes you as a spammer.

Please learn some modesty and learn to listen to others.

Quote:

Originally Posted by VBCurtis

If you'd stop calling your trivial observations "useful", you (and the forum) would be much better off.

JM Montolio A · 2018-02-27, 20:25

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

Batalov · 2018-02-27, 21:06

Quote:

Originally Posted by JM Montolio A

If one post is NOT NEW, please tell about it,
and can be deleted.
...

Delete, move.

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 nicknames in red is a moderator. You are talking to a moderator. A moderator who had enough dealing with spam.

JM Montolio A · 2018-02-28, 11:22

Reciba usted mis mejores grifos.

JMMA

Collag3n · 2018-02-28, 20:06

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

...

$1=\frac{3^5}{2^{11}}\cdot 7+\frac{3^4}{2^{11}}+\frac{3^3}{2^{10}}+\frac{3^2}{2^9}+\frac{3^1}{2^7}+\frac{3^0}{2^4}$

or rearranged:

$7=\frac{2^{11}}{3^5} - (\frac{2^7}{3^5}+\frac{2^4}{3^4}+\frac{2^2}{3^3}+\frac{2^1}{3^2}+\frac{2^0}{3^1})$

multiply by $3^5$ and you get

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

all you do is rearrange the whole thing.

2018-02-27, 12:28	#1
JM Montolio A Feb 2018 5·19 Posts	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

2018-02-28, 11:22	#10
JM Montolio A Feb 2018 5×19 Posts	Dr. Batalov, le agradezco su paciente comprensión. Reciba usted mis mejores grifos. JMMA Attached Thumbnails

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2018-02-27, 14:28	#2
CRGreathouse Aug 2006 5,987 Posts	Does this tell us anything more than "the number of steps is a function of n"?

2018-02-27, 15:33	#3
JM Montolio A Feb 2018 5·19 Posts	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: 3e+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) = n3bits(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

2018-02-27, 17:47	#5
JM Montolio A Feb 2018 5·19 Posts	Mire, Sr. Sergev Batalov. Replique a mis post ó no los conteste. Pero yo no entro aquí para las tonterias. De nadie. JM Montolio

2018-02-27, 18:11	#6
Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 10011101000101₂ Posts	Why do you 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?

2018-02-27, 20:25	#8
JM Montolio A Feb 2018 95₁₀ Posts	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

2018-02-28, 20:06	#11
Collag3n Feb 2018 2³ Posts	Nothing new. If you apply the Condensed Collatz function $\frac{3e+1}{2^g}$ starting from 7 until you reach 1, you get $11=\frac{3\cdot7+1}{2^1}$ $17=\frac{3\cdot(\frac{3\cdot7+1}{2^1})+1}{2^1}=\frac{3^2}{2^2}\cdot7+\frac{3^1}{2^2}+\frac{3^0}{2^1}$ ... $1=\frac{3^5}{2^{11}}\cdot 7+\frac{3^4}{2^{11}}+\frac{3^3}{2^{10}}+\frac{3^2}{2^9}+\frac{3^1}{2^7}+\frac{3^0}{2^4}$ or rearranged: $7=\frac{2^{11}}{3^5} - (\frac{2^7}{3^5}+\frac{2^4}{3^4}+\frac{2^2}{3^3}+\frac{2^1}{3^2}+\frac{2^0}{3^1})$ multiply by $3^5$ and you get $7\cdot 3^5=2^{11} - (2^7\cdot 3^0+2^4\cdot 3^1+2^2\cdot 3^2+2^1\cdot 3^3+2^0\cdot 3^4)$ or $7\cdot 3^5=2^{11} - (128\cdot 1+16\cdot 3+4\cdot 9+2\cdot 27+1\cdot 81)$ all you do is rearrange the whole thing.