Problem 58498. Compute the Sisyphus sequence
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I see now--because a(2)=3 you add p=5 instead of p=3, and then go back and use p=3 the next time. It seems like this is the only time this happens--from then on the next largest unused prime is always the next prime.
I still don't get why 3 is not "unused" in the alternate form on the second climb but is later on the third; it implies that it was "de-used". Is it that your step up can't be a doubling?
Yes, the flag only affects the addition of 3 and 5. After the two is used and a_n < the first prime, adding that prime produces an even number less than twice that prime. Half it, and it's smaller than that prime, hence smaller than the next prime as well. first four steps,
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