[seqfan] Re: A Mathar's conjecture

[P. Michael Hutchins]

(This has nothing to do with the content of Vladimir's post; jjust the
lexical level.)

Is space really at such a premium that everything has to be jammed together
like this?  Doesn't that make it much harder to read?

Cf.:

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.


On Fri, Nov 17, 2017 at 10:51 AM, Vladimir Shevelev <shevelev at bgu.ac.il>
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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> Seqfan Mailing list - http://list.seqfan.eu/
>