[seqfan] Re: A Mathar's conjecture

[israel at math.ubc.ca]

It is very easy to verify in Maple:

  > eq:=  eval((n+1)*a(n) +2*(-n-1)*a(n-1) +4*(-n+2)*a(n-2)
+8*(n-2)*a(n-3),  a = (n -> 2^(n-1)-binomial(n,floor(n/2)))) ;
  > simplify(convert(eq,factorial)) assuming n::even;

                         0

  > simplify(convert(eq,factorial)) assuming n::odd;

                         0

Cheers,
Robert

On Nov 17 2017, Vladimir Shevelev wrote:

>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
>
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>
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