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Nope. Just 2n: each chunk pair is added once without carry, and once won't carry=1.

For as long as radix=2, you either have a carry or you don't.


For a single addition, the radix is irrelevant, the carry is always zero or one: (r-1) + (r-1) = 2r - 2 < 2r.
But for a single addition you will also not save any time because you still need to do the select in series. The whole point of the radix 2^51 trick is that you can defer the normalization over several additions, but in order to do that your carry needs to be more than a single bit.

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