SIFT - why s+3 scales per octave?

Question

I have a problem with SIFT that I do not understand. Lowe [1] proposed in his work the s=3 levels of scale are enough for one octave. Afterwards, he mentioned that you need to compute s+3 levels.

Why there are 3 and not 2 additional levels required. I understand that you require one additional level above and one additional level below the scales since you search for extrema in neighbored scales.

For what is the third additional level of scale?

Thank you very much in advance!

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

Answer 1

We must produce s + 3 images in the stack of blurred images for each octave, so that final extrema detection covers a complete 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).

Answer 2

I can only find this is the paper:

As this graph shows, the highest repeatability is obtained when sampling 3 scales per octave, and this is the number of scale samples used for all other experiments throughout this paper. It might seem surprising that the repeatability does not continue to improve as more scales are sampled. The reason is that this results in many more local extrema being detected, but these extrema are on average less stable and therefore are less likely to be detected in the transformed image.

Where are you referring to when he says s+3?

I guess it also depends on the ratio of DOG filters you are using. I've seen DOGupper/DOGlower = 1.6 as it is pretty stable (D. Marr & E. Hildreth, "Theory of Edge Detection", 1980). This also approach a LOG filter.