[seqfan] A Mathar's conjecture

[Vladimir Shevelev]

Dear Seq Fans,

In 2012,  R. J. Mathar did conjecture for a(n)
= 2^(n-1) - binomial(n, floor(n/2)) (A093387):
 (n+1)*a(n) +2*(-n-1)*a(n-1) +4*(-n+2)*a(n-2)
+8*(n-2)*a(n-3)=0. 
Now I proved it. Firstly, let us prove that a(2*n+1)
= 2*a(2*n) + Catalan number which is
binomial(2*n, n)/(n+1) (A000108). Indeed, 2^(2*n)
- binomial(2*n+1, n) - 2*((2^(2*n-1) - binomial(2*n,n))
= 2*binomial(2*n, n) - binomial(2*n+1,n) =
binomial(2*n,n)*(2-(2*n+1)/(n+1)) = 
binomial(2*n,n)/(n+1) = A000108(n). Earlier
Emeric Deutsch noted that  a(2*n) = 2*a(2*n-1). 
Thus now we have a simple system of recursions.
Using them, we can prove Mathar's conjecture.
For example, let n be odd, n=2*m+1.
 By the left hand side of Mathar's conjecture, we
have (2*m+2)*a(2*m+1) - 2*(2*m+2)*a(2*m) -
 4*(2*m-1)*a(2*m-1) + 8(2*m-1)*a(2*m-2) =
(2*m+2)*(2*a(2*m) + A000108(m) - 2*a(2*m)) 
- 4*(2*m-1)*(2*a(2*m-2) + A000108(m-1) -
2*a(2*m-2)) = (2*m+2)*A000108(m) -
4*(2*m-1)*A000108(m-1) = 0. 
The latter equality is verified directly.

Best regards,
Vladimir