[seqfan] Re: Symmetric and Antisymmetric Linear Recurrences?

[Richard J. Mathar]

In partial answer to http://list.seqfan.eu/pipermail/seqfan/2014-September/013590.html :

The recurrence
  n=2: a(n)=4*a(n-1)-4*a(n-2)-4*a(n-3)+10*a(n-4)-4*a(n-5)-4*a(n-6)+4*a(n-7)-a(n-8)

has a "signature" (4,-4,-4,10,-4,-4,4,-1), if we define the signature
of a linear recurrence with constant coefficients as the coefficients in front
of a(n-1), a(n-2) etc on the right hand side.

If we consider sequences with rational generating functions with
denominators (1+x)^k*(1-x)^l, the sequences can be written as a polynomial
of order l in n plus (-1)^n times a polynomial of order k in n.

The following table shows explicitly the k>=1, l>=1 and signatures:
k=1, l=1, (0,-1,)
k=1, l=2, (1,1,-1,)
k=1, l=3, (-2,0,2,-1,)
k=1, l=4, (3,-2,-2,3,-1,)
k=1, l=5, (-4,5,0,-5,4,-1,)
k=1, l=6, (5,-9,5,5,-9,5,-1,)
k=2, l=1, (1,-1,-1,)
k=2, l=2, (0,2,0,-1,)
k=2, l=3, (-1,-2,2,1,-1,)
k=2, l=4, (2,1,-4,1,2,-1,)
k=2, l=5, (-3,1,5,-5,-1,3,-1,)
k=2, l=6, (4,-4,-4,10,-4,-4,4,-1,)
k=3, l=1, (2,0,-2,-1,)
k=3, l=2, (-1,2,2,-1,-1,)
k=3, l=3, (0,-3,0,3,0,-1,)
k=3, l=4, (1,3,-3,-3,3,1,-1,)
k=3, l=5, (-2,-2,6,0,-6,2,2,-1,)
k=3, l=6, (3,0,-8,6,6,-8,0,3,-1,)
k=4, l=1, (3,2,-2,-3,-1,)
k=4, l=2, (-2,1,4,1,-2,-1,)
k=4, l=3, (1,-3,-3,3,3,-1,-1,)
k=4, l=4, (0,4,0,-6,0,4,0,-1,)
k=4, l=5, (-1,-4,4,6,-6,-4,4,1,-1,)
k=4, l=6, (2,3,-8,-2,12,-2,-8,3,2,-1,)
k=5, l=1, (4,5,0,-5,-4,-1,)
k=5, l=2, (-3,-1,5,5,-1,-3,-1,)
k=5, l=3, (2,-2,-6,0,6,2,-2,-1,)
k=5, l=4, (-1,4,4,-6,-6,4,4,-1,-1,)
k=5, l=5, (0,-5,0,10,0,-10,0,5,0,-1,)
k=5, l=6, (1,5,-5,-10,10,10,-10,-5,5,1,-1,)
k=6, l=1, (5,9,5,-5,-9,-5,-1,)
k=6, l=2, (-4,-4,4,10,4,-4,-4,-1,)
k=6, l=3, (3,0,-8,-6,6,8,0,-3,-1,)
k=6, l=4, (-2,3,8,-2,-12,-2,8,3,-2,-1,)
k=6, l=5, (1,-5,-5,10,10,-10,-10,5,5,-1,-1,)
k=6, l=6, (0,6,0,-15,0,20,0,-15,0,6,0,-1,)

and we see that the signature of the 2nd row in the OP is (up to a global sign)
one of these, at k=6, l=2.
Apparently all the signatures in the class of (1+x)^k*(1-x)^l are 
palindromic, even or odd. (I'm too lazy to proof this in general.)

Note that this does not mean the inverse: palindromic linear recurrences
are not necessarily leading to denominators of the (1+x)^k*(1-x)^l form
of the g.f.'s.

Above table in Maple:
for k from 1 to 6 do
        for l from 1 to 6 do
                (1+x)^k*(1-x)^l ;
                f := expand(%) ;
                # print(k,l,f) ;
                printf("k=%d, l=%d, (",k,l) ;
                for d from degree(f)-1 to 0 by -1 do
                        printf("%d,",-coeff(f,x,d)) ;
                end do:
                printf(")\n") ;
        end do:
end do:

Richard J. Mathar