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A331440
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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.
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7
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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
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OFFSET
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1,2
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COMMENTS
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Theorem 1: Every positive numbers appears at least once.
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<j then S = 2^(j-i)*T, and T would have been a smaller choice for S. QED
So this is a permutation of the positive integers.
The sequence can naturally be seen as an irregular table where:
- T(n, k+1) = 2*T(n, k) for k = 1..A331442(n)-1.
(End)
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REFERENCES
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Eric Angelini, Posting to Math Fun Mailing List, Jan 04 2020.
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LINKS
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EXAMPLE
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The process begins like this:
Initially S = 1 is the smallest missing number, so we have:
S = 1, 2, 4, 8, 16, stop (because 16 contains S), S = 3, 6, 12, 24, 48, 96, 192, 384, stop, S = 5, 10, 20, 40, 80, 60, 320, 640, 1280, 2560, stop, S = 7, 14, 28, 56, 112, 224, 448, 896, 1792, stop, S = 9, 18, 36, 72, ...
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PROG
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(PARI) See Links section.
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CROSSREFS
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The inverse permutation is A331441. The lengths of the chains are given in A331442.
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KEYWORD
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nonn,base,tabf
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AUTHOR
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STATUS
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approved
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