[seqfan] Re: Conjecture about A127750

[Allan Wechsler]

Maybe I don't have to write and upload my proof because somebody cleverer
than I will beat me to it. Not sure what to hope for.

For me, at least, the reasoning went much more smoothly when I used
1-origin indexing, both in A127750 and in the two mutually inverse matrices.

I am making slow progress on writing up my proof. If somebody beats me to
it I won't mind!

On Thu, Feb 11, 2021 at 10:57 AM Richard J. Mathar <mathar at mpia-hd.mpg.de>
wrote:

> The recurrence of a row in A127749 is right-to-left
> A127749(n,n) = 2*n+1
> A127749(n,k) = -(2*k+1) *sum_{i=k+1... min(n,2*k)) A127749(n,i)/(2*i+1).,
> k<n.
> The recurrence means to "undo" the multiplication of the odd'ed index
> of the elements "right from" k, accumulate, and multiply back-in the
> (2*k+1).
> From this I'd conclude that A127749(n,k) = (2*k+1)*A111967(n,k),
> because A111967 is doing the same without the re-weighting by the
> odd integers. So the conjecture in A127750 is equivalent to the
> conjecture in A111967.
>
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