[seqfan] Re: Partitions and A101509

[Lior Manor]

Sorry for the unclear definition. Maximilian's definition is accurate. This
is what I meant. And as I assumed it is trivial to show that this is the
same as A101509. I will just add a comment mentioning that.

Lior

On Wed, Jan 14, 2009 at 10:18 PM, Maximilian Hasler <
maximilian.hasler at gmail.com> wrote:

> comparing to A120733 I suppose it is meant :
>
> a(n) = Number of matrices with positive integer entries such that sum
> of all entries is equal to n.
>
> n=1: [1]
> n=2: [2],[1,1], [1;1]
> ...
>
> The matrices must be of dimensions r x c <= n
> a(n) is the sum for k=1 to n of the product of
>
> sum of ways to write k as a product = number of divisors of k = tau(k)
>
> times
>
> the number of decompositions of n in k terms > 0 (respecting the order)
> = number of strictly increasing functions from [1,k] to [1,n] such that
> f(k)=n
> = number of ways to chose k-1 (different) numbers among n-1 (without order)
> = binomial(n-1,k-1)
>
> a(n) = sum( k=1,n, tau(k)*binomial(n-1,k-1))
>
> this is indeed equal to A101509.
>
> Maximilian
>
> > On Wed, Jan 14, 2009 at 4:04 PM,  <franktaw at netscape.net> wrote:
> >> I don't understand your definition.  Can you clarify; perhaps show the
> >> 3
> >> matrices you claim for n=2?
> >>
> >> Franklin T. Adams-Watters
> >>
> >> P.s. The word you want is "where", not "were".
> >>
> >> -----Original Message-----
> >> From: Lior Manor <lior.manor at gmail.com>
> >>
> >> Hi,
> >>
> >> Consider the number of possible partitions of n were the partitions are
> >> not
> >> a set of numbers but are elements of a matrix. Zeros are not allowed.
> >> (One
> >> variation were zeros are allowed is A120733).
> >>
> >> By manual calculation I get the first few elements to be: 1, 3, 7, 16,
> >> 35,
> >> 75, 159, 334
> >> This is the beginning of A101509.
> >>
> >> Is it trivial that these are the same sequences?
> >>
> >> Lior
> >>
> >>
> >> _______________________________________________
> >>
> >> Seqfan Mailing list - http://list.seqfan.eu/
> >>
> >
>



-- 
Lior
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