# SIFT: s + 3 intervals - resulting k

Based on another question in this subforum: SIFT - why s+3 scales per octave?

"For s=3 this means you will have s+3=6 blurred images (the Gaussian images shown in the paper in Figure 1 on the left). Having 6 Gaussian images will result in 5 DoG images (shown in Figure 1 on the right). This will allow you to do the extrema detection on s=3 scales (using the method shown in Figure 2)."

The above mentioned is pretty clear, now I was left wondering what the used sigma for the 6 images is? Let's say we go for 3 scales per ocatave like suggested, meaning we have to detect extrema points in 5 DoGs, making it necessary to smooth 6 images in the octave. According to the given formula $$k = 2^{1/s}$$, with $$s=3$$ we would need to more than double our initial sigma for 6 images due to $$\sigma_n = k^n \sigma_0$$, where $$n$$ is the number of images in one octave and $$\sigma_n$$ the used sigma for image $$n$$ of the octave. Since actually the first formula for $$k$$ is used to compute the number of intervals until $$\sigma_0$$ doubles, it feels odd. Is this intended or should the formula actually be $$k = 2^{1/(s+3)}$$.

[1] Distinctive Image Features from Scale-Invariant Keypoints D. G. Lowe Int. Journal of Computer Vision 60(2) (2004), pp. 91--110