[seqfan] Re: Formula for inverse Pascal's triangle A130595

[Anders Claesson]

Hi Dale

It follows from

T(n,k) = T(n-1,k-1) - T(n-1,k)

and n*T(n-1,k-1) = k*T(n,k) that

(n-k+1)*T(n-1,k-1) + (k-1)*T(n-1,k)
= n*T(n-1,k-1) - (k-1)*T(n-1,k-1) + (k-1)*T(n-1,k)
= n*T(n-1,k-1) - (k-1)*(T(n-1,k-1) - T(n-1,k))
= n*T(n-1,k-1) - (k-1)*T(n,k)
= n*T(n-1,k-1) - k*T(n,k) + T(n,k)
= T(n,k)

By the way, I think the o.g.f. expression for this entry, namely
1/(1+x+x*y), is wrong. It should be 1/(1+x-x*y)

Anders

On Sat, Nov 28, 2015, at 03:39 PM, Dale Gerdemann wrote:
> Dear Seqfans,
> 
> The following formula seems to work for A130595:
> 
> T(n,k) = (n-k+1)*T(n-1,k-1) + (k-1)*T(n-1,k)
> 
> Could someone explain why?
> 
> Dale
> 
> _______________________________________________
> 
> Seqfan Mailing list - http://list.seqfan.eu/