## triple structur of quadratic irreducible polynoms

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• A beautiful day i did an investigation of the triple structur of quadratic irreducible polynoms. Triple structur means for me that three quadratic polynoms
Message 1 of 27 , Oct 4, 2012
A beautiful day

i did an investigation of the triple structur of quadratic irreducible polynoms.

Triple structur means for me that three quadratic polynoms contain
all primes as divisor.

Detailed list under
http://109.90.219.147/devalco/basic_polynoms/triple.php

or because the ip could change
http://devalco.de 6.g)

The result is very amazing and gives hope that there migth be
a relationship concerning to the chemical periodic system:

180 of 440 polynoms occure in the triple structure

113 Polynome with f(x)=x^2+bx+c
63 Polynome with f(x)=2x^2+bx+c
4 Polynome with f(x)=3x^2+bx+c

if i am right there exist 112 elements which are stable.

the triple structur might have a relationship to the quarks,
which also appear as tripple.

If someone knows a better proof for the triple structur of the polynoms it would be very nice to send it to me.

By the way the investigation was made by looking for the first 64 primes starting with 3 up to 313

Mathematical feedback or improvements are welcome
Enjoy the prime sequences
Bernhard
• ... You are wrong. We know of only 80 elements with at least one stable isotope, namely those with atomic number Z
Message 2 of 27 , Oct 4, 2012
"bhelmes_1" <bhelmes@...> wrote:

> if i am right there exist 112 elements which are stable

You are wrong. We know of only 80 elements with at least one
stable isotope, namely those with atomic number Z < 83,
with the notable exceptions of Z = 43 (technetium) and
Z = 61 (promethium).

http://en.wikipedia.org/w/index.php?title=File:Isotopes_and_half-life.svg

Please do not set much store by the circumstance that 43, 61
and 83 are prime. (That they are even is more predictable.)

Moreover, the idea that a more or less arbitrary subset of
the infinitude of quadratic number fields might relate to
very messy nuclear physics, underlying this count of 80,
is improbable, in the extreme.

David
• ... Bernhard: I would consider carefully who this response is from (a very respected physicist) and give it substantial weight! (In fact, I would suggest
Message 3 of 27 , Oct 4, 2012
> > if i am right there exist 112 elements which are stable
> You are wrong. We know of only 80 elements with at least one stable isotope, namely those with atomic number Z < 83, with the notable exceptions of Z = 43 (technetium) and Z = 61 (promethium).
> http://en.wikipedia.org/w/index.php?title=File:Isotopes_and_half-life.svg
> Please do not set much store by the circumstance that 43, 61 and 83 are prime. (That they are even is more predictable.)
> Moreover, the idea that a more or less arbitrary subset of the infinitude of quadratic number fields might relate to very messy nuclear physics, underlying this count of 80, is improbable, in the extreme.
> David

Bernhard: I would consider carefully who this response is from (a very respected physicist) and give it substantial weight! (In fact, I would suggest giving up on any looking for such physical connection.) CC
• ... Since PrimeMogul has not yet ruled me off topic, I remark that, in nuclear physics, the prime Z = 83 is almost stable , since the half-life of bismuth 209
Message 4 of 27 , Oct 4, 2012
Chris Caldwell <caldwell@...> wrote:

>> Please do not set much store by the circumstance that
>> 43, 61 and 83 are prime.
> Bernhard: I would consider carefully who this response is from

Since PrimeMogul has not yet ruled me off topic, I remark that,
in nuclear physics, the prime Z = 83 is "almost stable",
since the half-life of bismuth 209 is about 2*10^19 years,
i.e. more than a billion times the age of the universe.
The most stable isotope for the prime Z = 43 has a
half-life of about 2 million years and the prime
Z = 61 has a corresponding half-life less than 6 years,
if I have read the tables aright.

David
• ... --well, I probably shouldn t be doing this but... 1. which elements are stable may be merely a matter of degree. For example it has been theorized (but
Message 5 of 27 , Oct 4, 2012
>
>
>
> "bhelmes_1" <bhelmes@> wrote:
>
> > if i am right there exist 112 elements which are stable
>
> You are wrong. We know of only 80 elements with at least one
> stable isotope, namely those with atomic number Z < 83,
> with the notable exceptions of Z = 43 (technetium) and
> Z = 61 (promethium).
>
> http://en.wikipedia.org/w/index.php?title=File:Isotopes_and_half-life.svg
>
> Please do not set much store by the circumstance that 43, 61
> and 83 are prime. (That they are even is more predictable.)
>
> Moreover, the idea that a more or less arbitrary subset of
> the infinitude of quadratic number fields might relate to
> very messy nuclear physics, underlying this count of 80,
> is improbable, in the extreme.
>
> David

--well, I probably shouldn't be doing this but...

1. which elements are stable may be merely a matter of degree. For example
it has been theorized (but not experimentally confirmed) that protons unstable, in which case presumably every element is unstable. I personally am 99.99999%
convinced every element is unstable, albeit perhaps with an extremely long halflife.
If so, we just haven't noticed this yet due to the long life.

2.the fact the unstable ones have Z=prime may actually mean something.
You tend to get more stability with even numbers due to spin pairing,
and one might imagine there are other phenomena which tend
to favor composite numbers. (For example, 3 quarks like to cohabit...)

3. For example in the atomic (rather than nuclear) world, the analogue
of the "most stable" nuclei would be the noble gas atoms, with
Z=2,10,18,36,54,86.
Every one is composite and even.

The atomic analogue of "least stable" would be the most-reactive atoms
namely the alkalis Z=3, 11, 19, 37, 55, 87
and the first two halides, Z=9, 17.
Of these, 5 out of 8 are primes (6 if you give credit to 9 for being a prime power).

I do not think these phenomena are entirely coincidental.
Compute the "null hypothesis probability" this all is due to a fluke for yourself.
I find over 99% confidence it is real, not a fluke.
That's not huge, but I'd bet on it.

4. But I don't recommend getting too excited by this. There's a lot more understanding
• ... Stable with respect to which reaction(s)? Yes, I know you re thinking of classical nuclear decay but as has been ... A possible mechanism is for the three
Message 6 of 27 , Oct 5, 2012
On Fri, 2012-10-05 at 04:49 +0000, warren_d_smith31 wrote:
>
>
>
> wrote:
> >
> >
> >
> > "bhelmes_1" <bhelmes@> wrote:
> >
> > > if i am right there exist 112 elements which are stable
> >
> > You are wrong. We know of only 80 elements with at least one
> > stable isotope, namely those with atomic number Z < 83,
> > with the notable exceptions of Z = 43 (technetium) and
> > Z = 61 (promethium).

Stable with respect to which reaction(s)?

Yes, I know you're thinking of classical nuclear decay but as has been
pointed out:

> it has been theorized (but not experimentally confirmed) that protons
> unstable, in which case presumably every element is unstable. I
> personally am 99.99999%
> convinced every element is unstable, albeit perhaps with an extremely
> long halflife.
> If so, we just haven't noticed this yet due to the long life.

A possible mechanism is for the three quarks in a nucleon to approach
each other closely enough to form a micro black hole which then
evaporates.

> 3. For example in the atomic (rather than nuclear) world, the analogue
> of the "most stable" nuclei would be the noble gas atoms, with
> Z=2,10,18,36,54,86.
> Every one is composite and even.
>
> The atomic analogue of "least stable" would be the most-reactive atoms
> namely the alkalis Z=3, 11, 19, 37, 55, 87
> and the first two halides, Z=9, 17.

Be very careful when making statements like this. David may be a
physicist but I'm a chemist by education I could make a convincing
argument that oxygen is more reactive than chlorine. I could even argue
that gold is more reactive than chlorine! (This one comes from my
background as a molecular spectroscopist. Last time I heard, gold
formed diatomic molecules with more other atoms than any element other
than oxygen. Not all chemistry occurs at room temperature ...)

Now we're seriously of topic. In a vague attempt to connect quantum
theory with prime numbers I'll advise you to search out "the spectrum of
Riemannium".

Paul
• ... Off list, Paul Leyland remarked on a well-known theoretical bug, affecting Wolfram, namely that for prime Z/2 = 37 there are 4 isotopes for which no decay
Message 7 of 27 , Oct 5, 2012

> in nuclear physics, the prime Z = 83 is "almost stable",
> since the half-life of bismuth 209 is about 2*10^19 years,
> i.e. more than a billion times the age of the universe.
> The most stable isotope for the prime Z = 43 has a
> half-life of about 2 million years and the prime
> Z = 61 has a corresponding half-life less than 6 years,
> if I have read the tables aright.

Off list, Paul Leyland remarked on a well-known
theoretical bug, affecting Wolfram, namely that for
prime Z/2 = 37 there are 4 isotopes for which no decay
has yet been observed, yet for which theory suggests that
alpha decay may occur:
http://en.wikipedia.org/wiki/Isotopes_of_tungsten

Meta-theory also suggests that all positive integers Z
might be unstable, since the proton, at Z = 1, might decay.

My position is strictly empirical (pace Paul and Warren).
I insist that with the notable exceptions of Z = 43 and
Z = 61 we have no evidence for decay at Z < 83.

David
• I am surprised with all these puerile associations that no one has yet pointed out that the Standard Model is currently based on just two prime numbers. 61
Message 8 of 27 , Oct 5, 2012
I am surprised with all these puerile associations that no one has yet pointed out that the Standard Model is currently based on just two prime numbers.

61 particles and forces, and 19 constants, and these primes are of course the higher of two prime pairs.

Try hard enough you can find unmeaningful meaning in anything.

On 5 Oct 2012, at 08:34, Paul Leyland <paul@...> wrote:

> On Fri, 2012-10-05 at 04:49 +0000, warren_d_smith31 wrote:
> >
> >
> >
> > wrote:
> > >
> > >
> > >
> > > --- In primenumbers@yahoogroups.com,
> > > "bhelmes_1" <bhelmes@> wrote:
> > >
> > > > if i am right there exist 112 elements which are stable
> > >
> > > You are wrong. We know of only 80 elements with at least one
> > > stable isotope, namely those with atomic number Z < 83,
> > > with the notable exceptions of Z = 43 (technetium) and
> > > Z = 61 (promethium).
>
> Stable with respect to which reaction(s)?
>
> Yes, I know you're thinking of classical nuclear decay but as has been
> pointed out:
>
> > it has been theorized (but not experimentally confirmed) that protons
> > unstable, in which case presumably every element is unstable. I
> > personally am 99.99999%
> > convinced every element is unstable, albeit perhaps with an extremely
> > long halflife.
> > If so, we just haven't noticed this yet due to the long life.
>
> A possible mechanism is for the three quarks in a nucleon to approach
> each other closely enough to form a micro black hole which then
> evaporates.
>
> > 3. For example in the atomic (rather than nuclear) world, the analogue
> > of the "most stable" nuclei would be the noble gas atoms, with
> > Z=2,10,18,36,54,86.
> > Every one is composite and even.
> >
> > The atomic analogue of "least stable" would be the most-reactive atoms
> > namely the alkalis Z=3, 11, 19, 37, 55, 87
> > and the first two halides, Z=9, 17.
>
> Be very careful when making statements like this. David may be a
> physicist but I'm a chemist by education I could make a convincing
> argument that oxygen is more reactive than chlorine. I could even argue
> that gold is more reactive than chlorine! (This one comes from my
> background as a molecular spectroscopist. Last time I heard, gold
> formed diatomic molecules with more other atoms than any element other
> than oxygen. Not all chemistry occurs at room temperature ...)
>
> Now we're seriously of topic. In a vague attempt to connect quantum
> theory with prime numbers I'll advise you to search out "the spectrum of
> Riemannium".
>
> Paul
>
>

[Non-text portions of this message have been removed]
• ... The constants (which are not) are all but fundamental; If the SM is based on 2 primes, then these are 2 & 3. (IMHO) But I fear there s something more...
Message 9 of 27 , Oct 5, 2012
On Fri, Oct 5, 2012 at 1:51 PM, <bobgillson@...> wrote:
> I am surprised with all these puerile associations that no one has yet pointed
> out that the Standard Model is currently based on just two prime numbers.
>
> 61 particles and forces, and 19 constants, and these primes are of course the higher of two prime pairs.

The "constants" (which are not) are all but fundamental;
If the SM is based on 2 primes, then these are 2 & 3. (IMHO)
But I fear there's something more...

Maximilian

> Try hard enough you can find unmeaningful meaning in anything.

Hm, I wonder what might be the "unmeaningful meaning" of these last
words of yours:

• Sent from my iPad ... Well, let s see - The Langrangian needs expanding by perhaps a further 10 constants to take into account the mass of neutrinos, which the
Message 10 of 27 , Oct 5, 2012

On 6 Oct 2012, at 02:16, Maximilian Hasler <maximilian.hasler@...> wrote:

> On Fri, Oct 5, 2012 at 1:51 PM, <bobgillson@...> wrote:
> > I am surprised with all these puerile associations that no one has yet pointed
> > out that the Standard Model is currently based on just two prime numbers.
> >
> > 61 particles and forces, and 19 constants, and these primes are of course the higher of two prime pairs.
>
> The "constants" (which are not) are all but fundamental;
> If the SM is based on 2 primes, then these are 2 & 3. (IMHO)
> But I fear there's something more...
>
> Maximilian
>
> > Try hard enough you can find unmeaningful meaning in anything.
>
> Hm, I wonder what might be the "unmeaningful meaning" of these last
> words of yours:
>
> > Sent from my iPad
>
Well, let's see - The Langrangian needs expanding by perhaps a further 10 constants to take into account the mass of neutrinos, which the current SM considers massless. Then we have 61 and 29. But that's a long shot and it would not tell us anything about primes. Hey ho!
>

[Non-text portions of this message have been removed]
• ... Z = 0 beta-decays with a half-life of a few minutes.
Message 11 of 27 , Oct 6, 2012
On Fri, 2012-10-05 at 17:14 +0000, djbroadhurst wrote:

> My position is strictly empirical (pace Paul and Warren).
> I insist that with the notable exceptions of Z = 43 and
> Z = 61 we have no evidence for decay at Z < 83.

Z = 0 beta-decays with a half-life of a few minutes.
• Dear David, ... The polynoms i have listed under http://109.90.219.147/devalco/basic_polynoms/ are not an arbitrary subset of the infinitude of quadratic
Message 12 of 27 , Oct 6, 2012
Dear David,

> Moreover, the idea that a more or less arbitrary subset of
> the infinitude of quadratic number fields might relate to
> very messy nuclear physics, underlying this count of 80,
> is improbable, in the extreme.

The polynoms i have listed under
http://109.90.219.147/devalco/basic_polynoms/
are not an "arbitrary subset of the infinitude of quadratic number fields"

I choose a subset of quadratic polynoms of the form ax^2+bx+c
which should fullfill some conditions:

1. f(0) should be equal 1 or a prime number
2. The polynomial should be irreducible, that means it should not be written as f(x)=(x+n)*(x+m), where n and m are integers.
3. Every polynomial constructs a special sequence of primes which is infinite and a different sequence from all other polynoms.

i found 440 different sequences to which i choose a polynom according
a special sort order:

1. I started the search for the polynoms from a=1 to a=4.
For each value of a i made a loop of c=1, -1, 2, -2, 3, -3, 5, -5 up to 100, -100.
For each value of a and b i made a loop for b from b=0, 1, -1, 2, -2, 3, -3, up to 300, -300.

2. I only choose sequences with b=0 mod a because of the special mathematical proof which is a limitation

3. The first appearance of a sequence was choosen, all sequences were sorted and by that proceeding the table was created.

4. The prime sequences are sorted by arising primes.

For these reasons i think the collection of sequences are not arbitrary as you described.

i checked up by a permutation of theses 440 basic polynoms
where a triple structure of the polynms occures and found 1193 different triples which i listed under
http://109.90.219.147/devalco/basic_polynoms/triple.php
( Sorry that this website is a little bit huge )

I will try to examine these tripples a little bit more,

The statement of 112 stable elements was of course wrong and
a mistake of my side.

If you have any constructiv suggestion how to examine these tripples,
i will try to examine these suggestion.

Best regards from the primes
Bernhard
• ... The limits of 4, 100 and 300 seem arbitrary to me. It is straightforward to used Pari-GP s isfundamental to identify the range of fundamental
Message 13 of 27 , Oct 6, 2012
"bhelmes_1" <bhelmes@...> wrote:

> http://109.90.219.147/devalco/basic_polynoms/
> are not an "arbitrary subset of the infinitude
>
> 1. I started the search for the polynoms from a=1 to a=4.
> For each value of a i made a loop of
> c=1, -1, 2, -2, 3, -3, 5, -5 up to 100, -100.
> For each value of a and b i made a loop for b from
> b=0, 1, -1, 2, -2, 3, -3, up to 300, -300.
> 2. I only choose sequences with b=0 mod a

The limits of 4, 100 and 300 seem arbitrary to me.
It is straightforward to used Pari-GP's "isfundamental" to
identify the range of fundamental discriminants, thereby
selected, and to identify those omitted from that range:

{arbitrary=[4,100,300];S=vector(2*prod(n=1,3,arbitrary[n]));N=0;
for(a=1,arbitrary[1],for(j=1,arbitrary[2],forstep(s=-1,1,2,c=s*j;
for(k=0,arbitrary[3]/a,b=k*a;F=factor(b^2-4*c);D=1;N++;
for(n=1,#F[,1],D*=F[n,1]^(F[n,2]%2));S[N]=if(D%4==1,D,4*D)))));
S=Set(S);T=vecsort(eval(S));R=[T[1],T[#S]];Found=#S-1;
Missed=sum(k=R[1],R[2],isfundamental(k))-Found;M=vector(Missed);
m=0;for(k=R[1],R[2],if(isfundamental(k)&&!setsearch(S,k),m++;M[m]=k));
M=vecsort(M);print(Found" fundamental discriminants were selected");
print("from the range "R" and "Missed" were omitted.");
print("First 10 omissions: ");print(vector(10,k,M[k]));}

24470 fundamental discriminants were selected
from the range [-399, 90392] and 3130 were omitted.
First 10 omissions:
[40805, 41208, 41212, 41613, 41617, 41621, 42024, 42028, 42429, 42433]

David
• ... instead of factor(b^2-4*a*c) Correcting my lamentable mistake, I obtained these stats: {arbitrary=[4,100,300];S=vector(2*prod(n=1,3,arbitrary[n]));N=0;
Message 14 of 27 , Oct 6, 2012

> factor(b^2-4*c)

Correcting my lamentable mistake, I obtained these stats:

{arbitrary=[4,100,300];S=vector(2*prod(n=1,3,arbitrary[n]));N=0;
for(a=1,arbitrary[1],for(j=1,arbitrary[2],forstep(s=-1,1,2,c=s*j;
for(k=0,arbitrary[3]/a,b=k*a;F=factor(b^2-4*a*c);D=1;N++;
for(n=1,#F[,1],D*=F[n,1]^(F[n,2]%2));S[N]=if(D%4==1,D,4*D)))));
S=Set(S);T=vecsort(eval(S));R=[T[1],T[#S]];M=[];P=vector(#S);m=0;
for(k=R[1],R[2],if(isfundamental(k)&&!setsearch(S,k),
if(k<1,M=concat(k,M),m++;P[m]=k)));
print(#S-1" fundamental discriminants were selected");
print("from the range "R" and "#M+m" were omitted.");
print("First 10 negative omissions: ");print(vector(10,k,M[k]));
print("First 10 positive omissions: ");print(vector(10,k,P[k]));}

25977 fundamental discriminants were selected
from the range [-1191, 91176] and 2098 were omitted.
First 10 negative omissions:
[-403, -407, -415, -419, -427, -431, -439, -443, -451, -455]
First 10 positive omissions:
[40805, 41613, 41617, 41621, 42433, 42437, 42445, 43253, 43261, 43265]

It seems to me to be oddly discriminatory for Berhard to
omit D = -13*31 = -403, while including D = -3*397 = -1191.

David
• Dear David, i did not understand your program. Please be so kind and publish the polynoms you have found for x^2+bx+c. I calculated the primes which appear as
Message 15 of 27 , Oct 7, 2012
Dear David,

i did not understand your program.

Please be so kind and publish the polynoms you have found for
x^2+bx+c.

I calculated the primes which appear as p | x^2+bx+c
sort them and choose a polynom which appears at first.

I think the polynoms are infinite but not the sequences they generate.

For the generation of the primes i used the following program:
http://109.90.219.147/devalco/basic_polynoms/pol.php?a=1&b=0&c=1#1
according to the polynom f(x)=x^2+1

I think polynoms like f(x)=ax^2+bx+c with a>4 could not generate
prime sequences according the algorithm because i get an estimation
p > 2x - 1

see http://109.90.219.147/devalco/basic_polynoms/pol.php?a=1&b=-4&c=1#3

Nice Greetings from the primes
Bernhard
• ... bhelmes_1 wrote: ... They can and do. For example, there is an infinite number of primes that divide 5*x^2 + 3*x + 1 for some integer x.
Message 16 of 27 , Oct 7, 2012
"bhelmes_1" <bhelmes@...> wrote:>

> I think polynoms like f(x)=ax^2+bx+c with a>4
> could not generate prime sequences

They can and do. For example, there is an infinite number
of primes that divide 5*x^2 + 3*x + 1 for some integer x.

Moreover it is easy to generate millions of those
primes, in sequence, along with a value of x for each.

All we care about is the discriminant
D = b^2 - 4*a*c = 3^2 - 4*5*1 = -11.

Then for every prime p for which -11 is
a square modulo p we have a solution to
5*x^2 + 3*x + 1 = 0 mod p.

Here are the first few primes,
each equipped with a value of x:

{forprime(p=2,137,if(kronecker(-11,p)>-1,
print([p,lift(polrootsmod(5*x^2+3*x+1,p))[1]])));}

[3, 1]
[5, 3]
[11, 3]
[23, 8]
[31, 4]
[37, 32]
[47, 5]
[53, 29]
[59, 10]
[67, 18]
[71, 11]
[89, 7]
[97, 25]
[103, 58]
[113, 47]
[137, 121]

It's just the same for any other fundamental discriminant.
That's why your insistence on tabulating polynomials
of arbitrarily chosen forms is both unnecessary and unhelpful.

All that matters is the fundamental discriminant,
which is an integer, D, defined so that D is squarefree
when D = 1 mod 4, or D/4 is squarefree, otherwise.

You can use Pari-GP to sieve out other discriminants:

> isfundamental(x): true(1) if x is a fundamental
> discriminant (including 1), false(0) if not.

David
• ... Yes, because there are an infinite number of fundamental discriminants. ... This is your very basic error. Every fundamental discriminant D produces its
Message 17 of 27 , Oct 7, 2012
"bhelmes_1" <bhelmes@...> wrote:

> I think the polynoms are infinite

Yes, because there are an infinite number
of fundamental discriminants.

> but not the sequences they generate.

This is your very basic error. Every
fundamental discriminant D produces
its unique infinite sequence of primes.

D = 1 produces all the primes.
Every other fundamental D produces half of the primes,
in the long run, and that sequence of primes is
unique to that value of D.

There are some values like D = -163 that appear
unusual if you look only at the first few primes,
but the asymptotic density is always half of the
primes, except of course for D = 1, when we
get all the primes.

We don't need tables that merely list the
first few kronecker values for an arbitrarily
selected subset of the infinitude of values of D.

We simply use a fast Euclidean algorithm
to compute kronecker(D,p), as in Pari-GP.

David
• ... When you have understood that you are merely listing triples of fundamental discriminants, you will see that the multiplicative property of the kronecker
Message 18 of 27 , Oct 7, 2012
"bhelmes_1" <bhelmes@...> wrote:

> http://109.90.219.147/devalco/basic_polynoms/triple.php
> I will try to examine these triples a little bit more

When you have understood that you are merely listing
triples of fundamental discriminants, you will see
that the multiplicative property of the kronecker symbol
guarantees that a triplet [D1, D2, D3] will generate all
the primes, in the manner of your tables, if the product
of a pair of fundamental discriminants, when reduced
to fundamental form, is equal to the third. In that case
it is impossible for all three of the kroneckers to be
negative, since (-1)*(-1) = +1.

A trivial example is the triple [12, -4, -3],

It is clear that the number of such triples is infinite.

You tables also show carelessness. For example
the final "triple" (ref 1193) should be removed
since it cannot generate any of these primes:

{forprime(p=2,2000,
if(kronecker([2364, 3832, 57],p)==-[1,1,1],
print1(p" ")));}

337 353 373 379 421 467 487 607 647 719 811 881 947 991 1013
1031 1039 1049 1061 1093 1151 1187 1259 1291 1487 1493 1499
1549 1559 1601 1613 1627 1693 1697 1747 1811 1951

There may be other such mistakes, which you should be able
to detect, simply using the kronecker symbol, as above.

David
• Well silly or not,       This could mean that every attempt we do over a schematic in Physics we approximate the reality &        therefore a
Message 19 of 27 , Nov 13, 2012
Well silly or not,

This could mean that every attempt we do over a schematic in Physics we approximate the "reality" &

therefore a structure of prime numbers which behold a certain degree of error related to the quantity

of calculus we put on the solution is possible!

Or I would say desirable.

Have you ever tought that 1,2,3... is a bit too lineare for physics?

________________________________
From: "bobgillson@..." <bobgillson@...>
To: Maximilian Hasler <maximilian.hasler@...>
Sent: Friday, October 5, 2012 11:13:36 PM
Subject: Re: [PrimeNumbers] nuclear & atomic physics & primes. (Warning: Kind of silly.)

On 6 Oct 2012, at 02:16, Maximilian Hasler <maximilian.hasler@...> wrote:

> On Fri, Oct 5, 2012 at 1:51 PM, <bobgillson@...> wrote:
> > I am surprised with all these puerile associations that no one has yet pointed
> > out that the Standard Model is currently based on just two prime numbers.
> >
> > 61 particles and forces, and 19 constants, and these primes are of course the higher of two prime pairs.
>
> The "constants" (which are not) are all but fundamental;
> If the SM is based on 2 primes, then these are 2 & 3. (IMHO)
> But I fear there's something more...
>
> Maximilian
>
> > Try hard enough you can find unmeaningful meaning in anything.
>
> Hm, I wonder what might be the "unmeaningful meaning" of these last
> words of yours:
>
> > Sent from my iPad
>
Well, let's see - The Langrangian needs expanding by perhaps a further 10 constants to take into account the mass of neutrinos, which the current SM considers massless. Then we have 61 and 29. But that's a long shot and it would not tell us anything about primes. Hey ho!
>

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