[seqfan] Re: The total number of parts in all partitions of n into consecutive parts.

Joerg Arndt arndt at jjj.de
Sat Jun 3 10:21:21 CEST 2017


* Omar E. Pol <info at polprimos.com> [Jun 02. 2017 09:28]:
> Dear Professors,
> 
> Is the total number of parts in all partitions of n into consecutive parts 
> equal to A204217(n)?

Yes:
  https://oeis.org/A001227 is (second comment, reworded)
"number of partitions of n into consecutive parts".
Now look at the g.f.
  Sum_{k>0} x^(k*(k+1)/2) / (1 - x^k).
[Ramanujan, 2nd notebook, p. 355.] - Michael Somos, Oct 25 2014

Th g.f. of https://oeis.org/A204217 is
  Sum_{n>=1} n * x^(n*(n+1)/2) / (1 - x^n).
Note the factor "n", corresponding to "number of all parts":
the g.f. can be understood to go along the Sylvester triangles
(the factor "x^(n*(n+1)/2)"), and the "n" is the number of parts
in it.


> Is also A204217 the row sums of the triangle A285914?

That may well be, but the sequence (and its sister, A237048) should
REALLY be cleaned up before I'd be ready to look at it.  In my
(strong) opinion all that "symmetric representation" voodoo should go.

The examples in both sequences suggest that they would much better be
defined via those partitions into consecutive parts, according to your
comment in A237048
"Conjecture 6: T(n,k) is also the number of partitions
 of n into exactly k consecutive parts."
It takes just a little thought to see that this is correct:
The first 1 in every column corresponds to the partition
  1+2+3+...+n (n parts) of n*(n+1)/2.
The 1's below (at n*(n+1)/2 + k, k \in N) correspond to
  (1+k)+(2+k)+(3+k)+...+(n+k) (n parts) of n*(n+1)/2 + n*k.
This construction gives all those partitions.

Best regards,   jj

> 
> Note that A204217 is defined as G.f.: Sum_{n>=1} n * x^(n*(n+1)/2) / (1 - 
> x^n).
> 
> https://oeis.org/A204217 
> https://oeis.org/A285914
> 
> Best regards.
> 
> Omar E. Pol
> 
> 
> --
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