a binomial sum.

[Paul D Hanna]

This sum can be restated as:

  a(n)=sum(m=0,n,(-1)^m*binomial(n+m+1,m+1))

seen more easily as the row sums of triangle:
1,
2,-3,
3,-6,10,
4,-10,20,-35,
5,-15,35,-70,126,
6,-21,56,-126,252,-462,
7,-28,84,-210,462,-924,1716,
8,-36,120,-330,792,-1716,3432,-6435,
9,-45,165,-495,1287,-3003,6435,-12870,24310,
10,-55,220,-715,2002,-5005,11440,-24310,48620,-92378,
11,-66,286,-1001,3003,-8008,19448,-43758,92378,-184756,352716,

The absolute sum of the rows gives: binomial(2*n+2,n+1)-1

so, subtracting twice the negative parts gives another sum:

  a(n)=binomial(2*n,n)-2*sum(m=0,n\2,binomial(n+2*m-1,2*m))+1

but I do not know of a simpler form, 
unless it involves the Catalan sequence 
and its' self-convolutions.

Paul


-- "pin" <pin at myway.com> wrote:


Hello seqfans!

I was about to submit the following signed sequence (offset 0),

1, -1, 7, -21, 81, -295, 1107, -4165, 15793, -60171, 230253, -884235, 3406105, -13154947, 50922987, -197519941, 767502945, -2987013067, 11641557717, -45429853651, 177490745985, ...,

but thought that I post here first requesting if someone is able to find a closed form of it.  Here is the sum that generates the sequence in Maple code.

a:=n->sum(sum(binomial(j-n-1,m),m=0..n),j=0..n);

Any ideas?

-Francois.

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