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 A331440 Let S = smallest missing positive number, adjoin S, 2*S, 4*S, 8*S, 16*S, ... to the sequence until reaching a term that has S as a substring; reset S to the smallest missing positive number, repeat. 7
 1, 2, 4, 8, 16, 3, 6, 12, 24, 48, 96, 192, 384, 5, 10, 20, 40, 80, 160, 320, 640, 1280, 2560, 7, 14, 28, 56, 112, 224, 448, 896, 1792, 9, 18, 36, 72, 144, 288, 576, 1152, 2304, 4608, 9216, 11, 22, 44, 88, 176, 352, 704, 1408, 2816, 5632, 11264, 13, 26, 52, 104, 208, 416 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Theorem 1: Every positive numbers appears at least once. Proof from Keith F. Lynch, Jan 04 2020: Since no nonzero power of 2 equals a power of 10, i.e., log(10)/log(2) is irrational, any sequence in which each term is the double of the previous term will start with every decimal number infinitely many times. So any S will terminate after a finite number of steps, and the next missing number will be used as S. QED Theorem 2: No term is repeated. Proof: Suppose N is repeated, so there are a pair of chains {S, 2*S, 4*S, ..., N = 2^i*S, ...}, {T, 2*T, 4*T, ..., N = 2^j*T, ...}, where T occurs after S. There are two cases. If i>=j then T = 2^(i-j)*S, so T was not a missing number. If i

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Last modified May 25 13:19 EDT 2022. Contains 354071 sequences. (Running on oeis4.)