We can assume I > B or better I -> infinity ( or n -> 0)

I'm not sure this is a safe assumption to make. It looks like B can be very large, e.g. if N = 10 and M = 100 then B = 64, and C = nB = 640, which is impossible since C is the number of coins cooling down, and cannot exceed N.

If we look at this iteratively, we start with N = nI coins. In the beginning, S = N = nI. Suppose n of those coins stake, meaning one UTXO stakes. Then S = N - n = nI - n = n(I - 1). If another n coins stake, then S = N - 2n = nI - 2n = n(I - 2). So if j is the number of staked blocks in the last 16 hours, then S = n(I - j). If I = j, i.e. if the number of staked blocks = number of UTXOs, this implies that all of the coins have staked and S = 0.

If it is the case that when you stake, all of the coins in a UTXO stake, then B should equal I, meaning that the maximum number of blocks you stake should equal the number of UTXOs you have.

If I understand your model correctly, it seems like there is an assumption that the probability of staking a block = number of your coins/number of total coins. Another way of saying this is if you have 10 coins in one UTXO out of 100, and your chance of staking is 10%, then someone else with 20 coins in one UTXO will have a 20% chance of staking. If this is correct, then by linearity of expectations, it shouldn't make a difference how many UTXOs you have - you could have one, or many, but the number of blocks you stake over a large enough time period should be the same. If however it is the case that you have a higher probability of staking a block by splitting coins in a single UTXO into two separate UTXOs, then an analysis can be made as to how much a person can exploit this.

Are you sure about: ...
As wallet =/= UTXO (or address =/= UTXO) a portion of a wallet will stake, but whole UTXO.

I'm not sure at all, I was wondering myself out of ignorance.

You can check list of UTXOs with a command

That's good to learn, thanks!