[snake_plissken]

Can someone explain why it goes from strictly multiplication/division for Somos 1,2 and 3 but then at Somos 4 the addition operation is added? It seems completely arbitrary.

[davesque]

Here's a definition from Wolfram math world that includes an addition operation for any valid value of k: https://mathworld.wolfram.com/SomosSequence.html

Are you talking about the behavior of the floor operator in determining the number of terms in the sum? For k in {1,2,3}, floor(k/2) = 1. So there would only be one term and thus no addition.

[ghusbands]

floor(1/2) is not 1.

[davesque]

Ah, heh. True. I guess Somos-1 would then be undefined? Because \sum_{n=1}^{0} doesn't make much sense.