[seqfan] Re: 4-ary words of length 2n

[israel at math.ubc.ca]

I see Richard Mathar has conjectured that A045952 satisfies 
n*a(n) +8*(3-4n)*a(n-1) +128*(2n-3)*a(n-2)=0.  Maple 16 confirms this:

> a:= n -> 2^(2*n-1) * binomial(2*n,n)+2^(4*n-1);
  Q:= n*a(n)+8*(3-4*n)*a(n-1)+128*(2*n-3)*a(n-2);
  convert(Q,GAMMA);

       n /n GAMMA(2 n + 1)   6 GAMMA(2 n - 1)   8 n GAMMA(2 n - 1)
  1/2 4  |---------------- + ---------------- - ------------------
         |             2                2                   2
         \ GAMMA(1 + n)         GAMMA(n)            GAMMA(n)

           16 n GAMMA(2 n - 3)   24 GAMMA(2 n - 3)\
         + ------------------- - -----------------|
                          2                    2  |
              GAMMA(n - 1)         GAMMA(n - 1)   /

> simplify(%);

              0

Robert Israel
University of British Columbia

On Aug 13 2012, David Scambler wrote:

>Seqfans,
>
> Take all 4-ary words of length 2n and split them into two subsets - those 
> where the combined count of 0's and 1's exceeds n and those where that is 
> not the case.
>
>Then these subsets are counted by A055787 and A045952 respectively.
>
>i.e. A055787 + A045952 = 2^4n.
>
> Not having the book from which these sequences derive I do not know 
> whether the book considered concrete examples such as this.
>
>dave
>https://oeis.org/A055787
>https://oeis.org/A045952
>
>
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