Re: Consecutive Numbers



rossum wrote:
On Sun, 14 Jun 2009 12:22:03 EDT, Maury Barbato
<mauriziobarbato@xxxxxxxx> wrote:

rossum wrote:

On Sat, 13 Jun 2009 14:02:36 EDT, Maury Barbato
<mauriziobarbato@xxxxxxxx> wrote:

Hello,
let k <= m <= n three positive integers.
Let S(k,m,n) be the number of subsets of
{1,...,n} with the following properties:

(1) S contains m elements

(2) S contains at least a sequence of at least k
consecutive numbers (that is, there exist j such
that j, j + 1, ..., j + k - 1 are in S)

I don't know the answer, except in the trivial
cases k = m or m = n. Do you have some idea?

Thank you for your attention.
My Best Regards,
Maury Barbato
Can you answer the question for S(1, m, n)?
Can you answer the question for S(2, m, n)?
Can you extrapolate from there?

rossum


Obviously, S(1,m,n) = (n choose m), but I don't
know how to find S(2,m,n). Some hint?
Build on what you have. You have an S with sequences 1 long. How can
you add numbers to ensure that you have sequences 1 longer that
before?

Perhaps aiming for some sort of inductive logic might help you here;
you want to move from S(x, n, m) to S(x+1, n, m).

rossum


Thank you for your attention.
Maury Barbato

MeAmI.org clarified:
3(m,n) = 2—»(n-^j,Miscellanea. 525. Then ?(m.n+l) = 2(n+l)+ m. S. 1..
2' r-3 ... r-3 tn-4 nt-4. S 2 r. S ( r-0 r-0 n«— 4 m— 4 r-0 r-0 l) + 2n
(2—»-l)-(m- agreeing with (4). ...http://biomet.oxfordjournals.org/cgi/
reprint/44/3-4/524.pdf
.

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