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

[Omar E. Pol]

Joerg,

Thank you very much for the explanation. 
I will edit those sequences. 

Best regards
Omar

..............
* 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)?
>
>Joerg:
>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.
>
>Omar:
> Is also A204217 the row sums of the triangle A285914?
>
>Joerg:
>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
>