I read this as -2, 9, -1, 6, 0, 5, 1, 6, 2, 9 (inserting apparently missing commas).
Note that 9+(-2+(-1))=6; 6+(-1+0)=5; 5+(0+1)=6; 6+(1+2)=9.
This can be written in a general form: a[2n]+a[2n-1]+a[2n+1]=a[2n+2] where a[i] means a sub i and n is a positive integer (n≥1). Example: n=1: a4=a2+a1+a3=9-2-1=6; n=2: a6=a4+a3+a5=6-1+0=5.
To generate the series we need to start with a1=-2, a2=9, a3=-1. These are the generators. But we also see that a3=a1+1. So in the general case we have a[2n+1]=a[2n-1]+1. There is an explicit relationship between odd terms: a[2n-1]=n-3. The only generator is a2=9 because the other terms can be derived from n alone.
a[2n]=9 when n=1; a[2n+2]=a[2n]+n-3+n-2=a[2n]+2n-5. When n=1, a4=a2-3=9-3=6; when n=2, a6=a4-1=6-1=5, when n=3, a8=a6+1=5+1=6.
Inspecting this more closely and focussing on the a[2n] terms, a4=9+(2*1-5); a6={9+(2*1-5)}+(2*2-5); a8={{9+(2*1-5)}+(2*2-5)}+(2*3-5); a10={{{9+(2*1-5)}+(2*2-5)}+(2*3-5)}+(2*4-5).
a10=9+2(1+2+3+4)-4*5=9+20-20=9.
The sum of natural numbers from 1 to n is given by n(n+1)/2.
So we have a pattern for a[2n+2]=9+2n(n+1)/2-5n=9+n^2+n-5n=n^2-4n+9. This can be written as:
a[2n]=(n-1)^2-4(n-1)+9=n^2-2n+1-4n+4+9=n^2-6n+14.
And the pattern for a[2n-1]=n-3.
These two formulae enable us to work out any term, including beyond the 10th term, assuming the pattern continues.