FW: another Motzkin triangle, no shortage of'm

[Ralf Stephan]

Wouter Meeussen: 
> > {1, 1, 2, 4, 9, 21, 51, 126, 315, 792, 1998, 5049, 12771, 32319, 81810, 207117, 524394}

This seems to be 1/3*(A062882(n)+A062882(n+1)) or 1/9 the partial
sums of A062882. Working out a(n+2)-a(n+1)-a(n) gives A090400/3 so
the g.f. should have as denominator (1-x-x^2)*(1-3x+3x^3).

> > {1, 1, 2, 4, 9, 21, 51, 127, 322, 826, 2135, 5545, 14445, 37701, 98514, 257608, 673933}

This one appears to be A033192(n)+A033192(n+1) = C(F(n)+1,2)+C(F(n+1)+1,2).
Or 1/2*(A059727(n)+A059727(n+1)), so the g.f. is
[1-2x-x^2+x^3]/[(1-3x+x^2)(1-x-x^2)] but you have to mul. with x and add 1.

As these are all rational gfs, superseeker should be able to give you 
better answers.

ralf