## Re: Greatest Prime Factor Sets?

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• Hello again. It has taken me approximately a week to analyze and rigourosly formulate this fundamental truth: the default address for replies in Yahoo groups
Message 1 of 7 , Feb 2, 2007
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Hello again.

It has taken me approximately a week to analyze and rigourosly
formulate this fundamental truth: the default address for replies in
Yahoo groups is the author of the message.

Before I made this independent discovery, I sent two messages to
myself and one to Payam Samidoost, who kindly and brilliantly replied.

So this is actually my third response to Phil and Payam., whom I thank
(for the third time!) for their useful and enlightening help.

I guess I should explicitly state what I've worked out so far so my
questions aren't so cryptic. First, these sets are a way of mapping
the primes to the integers; second, they have a very simple iterative
structure, and because of this there is a unique, finite set of
integers associated with each prime.

1) Mapping:. every integer can be uniquely identified by the prime
that is its greatest (or the prime that is its least) prime factor and
its "position" in these sets. That is, if all the integers whose
greatest prime factor is 3 are arranged (for instance) in ascending
order, 3 is first, 6 is second and so one. In general, there is a
total function f(p,k)=x where S is the set of all integers with
greatest prime factor p and the integers in S are ordered such that x
is the kth integer in S.

My questions are: what is it called when you map the integers to the
primes using factorizations? Are there many ways of doing it? What
is this way called? (It's all very well to stumble over these things,
but trying to track them down isn't working out for me.)

2) I have the same two question about these sets:

{2}
{4, 3}
{8, 6, 9, 5}
{16, 12, 18, 27, 10, 15, 25, 7}

Among their characteristics is that, where P(n) is the nth prime
number, designating 2 as the first prime, each set contains 2^(n-1)
integers, including P(n) and 2^n.

Here's where they come from: each "greatest prime factor set" can be
divided into subsets based on the number of "not necessarily distinct
prime numbers" in the factorization.

(The symbol for that is a big omega. I'll call it G(m) where m is an
integer.)

For the greatest prime factor set of P(n), the number of integers
Where G(m)=r is C(n+r-1,n) i.e., n choose k with repetitions.

(I won't work that out here, it's pretty straightforward.)

There are several ways to show the next step, but here's the coolest
one: a table using the combination formula above:

n=1 n=2 n=3 n=4
r=1 1 1 1 1
r=2 1 2 3 4
r=3 1 3 6 10
r=4 1 4 10 20

Each number represents how many integers are in the subset. The
finite sets are based on the diagonals. To show it with the actual
integers,

n=1 n=2 n=3
r=1 2 3 5
r=2 4 6, 9 10, 15, 25
r=3 16 12, 18, 27 20, 30, 45. 50, 75, 125

Well, I'm a rank amateur and I'm very proud of having worked that out
myself, but what do you call it? It feels kind of deep.
• Hi Phil ... I must have at some point looked at the Wolfram article on least prime factor but I somehow missed the connection. Wolfram says that lpf(n) is
Message 2 of 7 , Feb 3, 2007
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Hi Phil

>The obverse, those which have the same least prime factor, are used
> extensively.

I must have at some point looked at the Wolfram article on "least
prime factor" but I somehow missed the connection. Wolfram says
that lpf(n) is standard, so I guess I can use gpf(n).

There are two properties that "gpf" sets have that interest me, and
"lpf" sets have only one of them, but it took me a while to see that
it doesn't have the other.

The property they share is that each integer, by definition, has one
and only one least- or greatest- prime factor, so you get the mapping.

The property they don't share is the one that leads to the finite sets
(although it may be there and I'm not seeing it). This is because
using the gpf puts an upper limit on the number of integers with the
same "not necessarily distinct factors".

lpf gpf

3 3
3*2 3*2*2 ... 2*3 3*3
3*3 3*3*3 ... 2*2*3 2*3*3 3*3*3
3*5 ...

lpf sets are n-dimensional, so to speak. gpf sets are two dimensional.

>
> Just give it an abstract symbolic name, and a tight definition, and
you'll be
> fine. Something like:
>
> H_p := { n : p|n /\ (q|n->q<=p) }

For my purposes, it's useful to define everything in terms of the n in
P(n), as defined above, with P(1)=2. So I don't use n for anything
else, and I was using S_n for the sets. (Except I either wrote it in
pencil or used word, so I got to use subscripts and symbols and stuff.
I never tried to do all this in text until I joined this group.)

I can't think of a way to define P(n) without using words, so my
(latest) inelegant definition was:

Where P(n) is the nth prime number, S_n={x: gpf(x)=P(n)}

(Except I wrote out "P(n) is the greatest prime factor of x" until you
guided me to lpf sets.)

Another way of defining gpf sets is "all possible products of a prime
with every possible positive integer exponent of every prime less than
or equal to itself."

I don't know how to say "every possible positive integer exponent"
symbolically.

Sorry, I am a wordy guy and I write long posts. I am very grateful
• ... p-smooth sets are infinite for all p. Therefore so are sets of numbers with the same gpf. ... Countable and countable. I m not sure I see the difference.
Message 3 of 7 , Feb 3, 2007
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--- "joel.levenson" <joel.levenson@...> wrote:
> Hi Phil
>
> >The obverse, those which have the same least prime factor, are used
> > extensively.
>
> I must have at some point looked at the Wolfram article on "least
> prime factor" but I somehow missed the connection. Wolfram says
> that lpf(n) is standard, so I guess I can use gpf(n).
>
> There are two properties that "gpf" sets have that interest me, and
> "lpf" sets have only one of them, but it took me a while to see that
> it doesn't have the other.
>
> The property they share is that each integer, by definition, has one
> and only one least- or greatest- prime factor, so you get the mapping.
>
> The property they don't share is the one that leads to the finite sets
> (although it may be there and I'm not seeing it). This is because
> using the gpf puts an upper limit on the number of integers with the
> same "not necessarily distinct factors".

p-smooth sets are infinite for all p. Therefore so are sets of numbers with the
same gpf.

> lpf gpf
>
> 3 3
> 3*2 3*2*2 ... 2*3 3*3
> 3*3 3*3*3 ... 2*2*3 2*3*3 3*3*3
> 3*5 ...
>
> lpf sets are n-dimensional, so to speak. gpf sets are two dimensional.

Countable and countable. I'm not sure I see the difference.

Dense and not dense is the distinction that you could draw. P-smooth numbers
are not dense in the integers, so neither are sets sharing gpfs.

Phil

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• Hi Phil ... numbers with the same gpf. gpf sets are infinite, and their union is the set of all integers greater than one, and their intersection is empty.
Message 4 of 7 , Feb 3, 2007
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Hi Phil

> p-smooth sets are infinite for all p. Therefore so are sets of
numbers with the same gpf.

gpf sets are infinite, and their union is the set of all integers
greater than one, and their intersection is empty.

What interests me about organizing the integers in terms of their gpfs
is that each one can be uniquely identified by its gpf and its
position among the rest of the integers with that gpf (when they are
placed in ascending order, for instance.) But a p-smooth number is
also p-smooth for every prime greater or equal to p, so it isn't
uniquely identified. (Payam suggested calling numbers with the same
gpf "strictly p-smooth numbers" But, I think, from my point of view,
it is simpler to think in terms of gpfs. Then you don't have to bring
in any concepts beyond the fundamental theorem of arithmetic and the
definition of "greatest.")

I don't know if mapping the primes to the integers is interesting or
useful, or how many ways it can be done, but this way caught my
attention.

> > lpf gpf
> >
> > 3 3
> > 3*2 3*2*2 ... 2*3 3*3
> > 3*3 3*3*3 ... 2*2*3 2*3*3 3*3*3
> > 3*5 ...
> > >
>> lpf sets are n-dimensional, so to speak. gpf sets are two dimensional.

> > Countable and countable. I'm not sure I see the difference.
> > Dense and not dense is the distinction that you could draw.
> > P-smooth numbers are not dense in the integers, so neither
> > are sets sharing gpfs.

(I was using n-dimensional sort of whimsically. I just meant that if
you try to write out all the lpf numbers on a piece of paper you have
to go off in all directions.)

What I am interested in here is the combinatorial structure, involving
subsets of gpf sets, defined by the number of "not necessarily
distinct primes." There are an infinite number of these subsets,
their union is the gpf set, and their intersection is the null set.
But each one is finite, and the formula for how many elements there
are in each set, where P(n) is the nth prime and r is the number of
not necessarily distinct prime factors, is "n choose r with
repetitions" which is C(n+r-1, n). So you can get a very simple
structure associating the integers with the primes, and with a little
manipulation you get these finite sets:

{2}
(4,3}
{8, 6, 9, 5}
(16, 12, 18, 27, 10, 15, 25, 7}

This pattern goes on forever. Among other things, the nth set has
2^(n-1) and contains P(n) and 2^n. The pattern of the other numbers
has to do with the "not necessarily distinct prime factor" subsets.
(My second post has enough detail to explain it adequately, I think.)

I was just surprised to find so much structure, and have been
exploring it as best I can.

Thanks for your help! And your time. I'm sorry, I can't help being
so wordy.
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