- Hi Phil

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

I must have at some point looked at the Wolfram article on "least

> extensively.

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.

>

you'll be

> Just give it an abstract symbolic name, and a tight definition, and

> fine. Something like:

For my purposes, it's useful to define everything in terms of the n in

>

> H_p := { n : p|n /\ (q|n->q<=p) }

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

for your and Payam's help. - --- "joel.levenson" <joel.levenson@...> wrote:
> Hi Phil

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

>

> >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".

same gpf.

> lpf gpf

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

>

> 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.

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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Ask your question on www.Answers.yahoo.com - 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

(I was using n-dimensional sort of whimsically. I just meant that if

> >

> > 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.

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.