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


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