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Primes>2are all odd, so in the former sequence every ascents will be immediately followed by at least one descent. But squares alternate parity, so in this sequence, ascents (after the first) will always occur in pairs, followed by at least one descent. (End)

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Allan C. Wechsler's initial remarks:(startStart):

But later, Gareth McCaughan did the "naive" probabilistic calculation more carefully, and I am now convinced that he and Lynch are correct and that the sequence eventually permanently exceeds any given value, most often without everequallingequalingthat value exactly.

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This variation onA350877was proposed byKeith F. Lynchon 2022 January 28. Because the former sequence uses prime increments, and because squares grow much faster than primes, Lynch felt that my conjecture about the former sequence (that all integers eventually appear) would not be true of this sequence. I responded that because squares nevertheless grow much more slowly than exponentials, we should still expect the same basic argument, that descents to the vicinity of 1 should happen on the order of once per "era", to apply, and I stillthinkthoughtthat all valueswillwouldappear, albeit perhaps very slowly.

But later, Gareth McCaughan did the "naive" probabilistic calculation more carefully, and I am now convinced that he and Lynch are correct and that the sequence eventually permanently exceeds any given value, most often without ever equalling that value exactly.

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Hans Havermann, <a href= "/A350742/a350742_1.png ">A350742minima</a> in the first 150 billion terms.

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