Re: Possible results from three variables

On 13 Jul, 21:18, quasi <qu...@xxxxxxxx> wrote:
On 13 Jul 2007 12:11:58 -0700, "sttscitr...@xxxxxxxxx"

<sttscitr...@xxxxxxxxx> wrote:
On 13 Jul, 19:48, quasi <qu...@xxxxxxxx> wrote:
On Fri, 13 Jul 2007 03:22:59 -0700, "sttscitr...@xxxxxxxxx"

<sttscitr...@xxxxxxxxx> wrote:
On 12 Jul, 14:57, quasi <qu...@xxxxxxxx> wrote:
On Thu, 12 Jul 2007 07:25:37 -0000, "bak...@xxxxxxxxx"

<bak...@xxxxxxxxx> wrote:
Given three variables, say x, y, z with each variable being an integer
from 1 to 10, how many possible values are there from the equation of
x * y * z? The quickest and incorrect answer is 1000 (from 10x10x10).
This is wrong because some of the combinations actually the same, e.g.
1*2*4 = 1*4*2 = 4*1*2 = 1*8*1. Doing 10C3 is also wrong.

Is there any formula that we can use to find out the number of
possibilities for the different integer range (e.g. 1 - 5 or 1 - 8)?

For positive integers n,k, let f(n,k) be the number of possible
products of k integers from the range 1 to n inclusive.

It's obvious that f(1,k)=1 for all positive integers k.

It's also obvious that f(n,1)=n for all positive integers n,

For a fixed positive integer n, it appears that f(n,k) is always a
polynomial in k.

However, for a fixed positive integer k, with the exception of k=1, it
appears that f(n,k) is _not_ a polynomial in n. This seems somewhat
mysterious.

Here are some of the formulas for f(n,k), for n from 1 to 10:

f(1,k) = 1
f(2,k) = k + 1
f(3,k) = ((k + 1)(k + 2))/2
f(4,k) = (k + 1)^2
f(5,k) = ((k + 1)(k + 2)(2 k + 3))/6
f(6,k) = ((k + 1)^2 (k + 2))/2
f(7,k) = ((k + 1)(k + 2)(k + 3)(3 k + 4))/24
f(8,k) = ((k + 1)^2 (k + 2)(k + 3))/6
f(9,k) = ((k + 1)^2 (k + 2)^2)/4
f(10,k) = ((k + 1)^2 (k + 2) (2 k + 3))/6

I was wondering how you are deriving your f(m,k).

Brute force count followed by interpolation.

In the case of f(5,k) ,say, would you be associating
1,2,3,4,5 with the polynomial ((1 +w+w^2) + x+y)^k
=(P+Q+R)^k and then using various relations
such as P^m =2m+1, Q^n = 1, etc.
to determine the number of distinct terms in the expansion
=(P+Q+R)^k = P^k + Q^k +R^k + P^(k-1)Q + P^(k-1)R + ?

I thought about this idea, but didn't try it.

It would at least explain why you always obtain polynomials.
The final result always involves sums that are functions
of k and certain binomial coefficients that are functions of k.

So can you then prove what I called Conjecture (4)?

I have no personal interest in proving Conjecture (4).
But if you follow through some examples, there us no
"mystery" as to why f(m,k) are polynomials.


.

Relevant Pages

  • Re: Possible results from three variables
    ... "mystery" as to why fare polynomials. ... No, you haven't unraveled the mystery, not from my point of view. ... It's easy since 1,2,3 are all not composite. ... such triples, yielding ...
    (sci.math)
  • Re: Possible results from three variables
    ... "mystery" as to why fare polynomials. ... No, you haven't unraveled the mystery, not from my point of view. ... Are you implying that, based on your outlined strategy, the proof that ... It's easy since 1,2,3 are all not composite. ...
    (sci.math)
  • Re: Possible results from three variables
    ... to determine the number of distinct terms in the expansion ... whether the counting strategy you outlined _already_ proves it. ... "mystery" as to why fare polynomials. ... No, you haven't unraveled the mystery, not from my point of view. ...
    (sci.math)
  • Re: Possible results from three variables
    ... I wasn't challenging you to prove it. ... whether the counting strategy you outlined _already_ proves it. ... "mystery" as to why fare polynomials. ... No, you haven't unraveled the mystery, not from my point of view. ...
    (sci.math)
  • Re: Possible results from three variables
    ... whether the counting strategy you outlined _already_ proves it. ... "mystery" as to why fare polynomials. ... No, you haven't unraveled the mystery, not from my point of view. ... Are you implying that, based on your outlined strategy, the proof that ...
    (sci.math)