Trying to re-write Simpson's Rule: mistake? Pre-Question (edited): Thanks Arthur 
Orignal Problem: The standard form of Simpson's Rule requires an even value of n so that you can make a series of parabolas
Parabola 1 has area $$\frac{x_n-x_0}{n} * \frac{f(x_0) + 4f(x_1) + f(x_2)}{3}$$
Parabola 2 has area $$\frac{x_n-x_0}{n} * \frac{f(x_2) + 4f(x_3) + f(x_4)}{3}$$
…
And the entire sum becomes $$\frac{x_n-x_0}{n} * \frac{f(x_0) + 4f(x_1) + 2f(x_2) + … + 2f(x_{n-2}) + 4f(x_{n-1}) + f(x_n)}{3}$$
I wanted to see what this would look like if extra parabolas, these graphed with points of odd-even-odd values, were added to the existing series of even-odd-even parabolas.
Googling "overlapping Simpson's Rule," "Simpson's Rule variants," and the such only gave me A) a very messy combination of the standard and the 3/8 Rules, and B) the even higher order NC quadratures that apparently lose accuracy to a "Runge's phenomenon."
I tried to figure out a simpler one myself by starting with the series of f values:
$$f(x_0) + 4f(x_1) + 2f(x_2) + … + 2f(x_{n-2}) + 4f(x_{n-1}) + f(x_n)$$
=>
$$f(x_0) + 5f(x_1) + 6f(x_2) + … + 6f(x_{n-2}) + 5f(x_{n-1}) + f(x_n)$$
and then seeing how the $\frac{x_n-x_0}{n}$ term would need to change in response.
I figured that since there were now $n-1$ parabolas instead of the original $\frac{n}{2}$, and the new parabola would each have roughly the same area as the 2 originals nearby, I should multiply my new result by $\frac{n/2}{n-1}$ to account for the increased number of areas going into the sum.
$\frac{x_n-x_0}{n}$ now became $\frac{x_n-x_0}{2(n-1)}$, and I went to Excel to test this formula with $f(x) = cosx + 1$ with 10 and 100 subintervals from 0 to 3
(correct area: $_0∫^3 (cosx+1)dx = (sin3+3)-(sin0+0)≈3.14112$.)
At n=10, the standard 1424…4241 formula gave a relative error of 2*10-6 and my new 1566…6651 gave a relative error of 3*10-3.
At n=100, the "142" version gave a relative error of 2*10-10 and my "156" version gave a relative error of 4*10-4
Is this formula "correct" (but still useless, and I should stick to the established formulae), or did I make a mistake somewhere that I can fix to make mine more accurate?
 A: You did not calibrate the formula correctly, introducing extra multiplicative error. The factor in front of the sum should be such that the formula is exact for the constant function $f(x)\equiv 1$. Writing $n=2k$, we see that the sum 
$$f(x_0) + 5f(x_1) + 6f(x_2) + … + 6f(x_{n-2}) + 5f(x_{n-1}) + f(x_n)$$
evaluates to 
$$2+5k+6(k-1) = 11k-4 = \frac{11n-8}{2}$$
Since $\int_a^b 1\,dx = (b-a)$, the correct formula is 
$$\int_a^b f(x)\,dx \approx \frac{2(b-a)}{11n-8}\left[ f(x_0) + 5f(x_1) + 6f(x_2) + \dots + 6f(x_{n-2}) + 5f(x_{n-1}) + f(x_n)\right]$$
For $\int_0^3 (\cos x+1)\,dx$ with $n=10$ this gives $3.149554$, which is actually worse than your version. I think you got lucky there because using too small factor in front compensated for the fact that the formula overestimates. 
So: your formula has error $O(n^{-1})$ (seen by testing on constant function); the corrected version above has error $O(n^{-2})$  (seen by testing on $f(x)=x^2$); both are far inferior to Simpson's rule. 
Rule of thumb: if your formula is not exact for quadratic polynomials, it will perform worse than Simpson's rule. 
