right-half-sum triangle

[Ralf Stephan]

>   1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1
>   .  .  .  .  .  .  .  .  1  2  3  4  5  6  7  8
>   .  .  .  .  .  .  .  .  .  .  .  .  5 11 18 26
>   .  .  .  .  .  .  .  .  .  .  .  .  .  . 18 44
>   .  .  .  .  .  .  .  .  .  .  .  .  .  .  . 44
> 
> Therefore, a(n)=44.
> 
> I'm mostly interested in a closed formula for this sequence.

Let k be the row number.
Although the rows at first look simple (they are polynomials of degree k),
they have changing start values dependent on the previous row polynomial
which itself depends on its start value:

s(1,n)=1
s(2,n)=1
s(3,n)=2^(n-2)+1
s(4,n)=a k-degree polynomial function of s(3,n)
...

I guess a(n) = n-degree polynomial function of s(n,n) has the
very general form p1(2^p2(2^p3(...pn()))) where the ps are polynomials
of decreasing degree of n and the power tower has altitude n.

Thus a 'closed form' in the sense of Zeilberger seems impossible.
But maybe a relatively simple nonlinear recurrence?


ralf

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