[seqfan] Re: A question about partitions

[bacher]

Here a proof:

Write n=1+2+..+(a-1)+b
with a<=b<=2a.

Then a is maximal number of distinct parts of
a partition of $n$.

Any integer  j in {1,2,..,a}
defines now either a partition of n into consecutive parts or
a partition of n with (largest part -smallest part)=number of parts
(such partitions have at last 2 parts and are obtained by considering
consecutive parts, except for a missing hole).

Indeed, if j divides (n-(1+2+..+j)=n-{j+1\choose 2},
we can consider q+1,q+2,q+3,.., q+j where q is the quotient.

Otherwise, we have a rest r when dividing (n-{j+1\choose 2}) by j
and we add 1 to the r largest parts of
q+1,q+2,.. q+j where qj+r=n-{j+1\choose 2}.

Best wishes, Roland


Neil Sloane <njasloane at gmail.com> a écrit :

> There is an open question, and probably not too difficult:
> :
> Show that A296509 = A238005.
>
> The LHS is   (maximal number of parts in any partition of n into distinct
> parts) minus (number of ways to partition n into consecutive parts)., which
> is  simply A003056(n) - A001227(n), and both of these are well-studied
> classics
>
> The RHS is Number of strict partitions of n such that (greatest part) -
> (least part) = (number of parts).
> Strict partitions means no repeated parts.
>
>
>
> Best regards
> Neil
>
> Neil J. A. Sloane, Chairman, OEIS Foundation.
> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
> Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
> Phone: 732 828 6098; home page: http://NeilSloane.com
> Email: njasloane at gmail.com
>
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