A331441 -id:A331441 - cp's OEIS frontend

A331440 Let S = smallest missing positive number, adjoin S, 2S, 4S, 8S, 16S, ... to the sequence until reaching a term that has S as a substring; reset S to the smallest missing positive number, repeat.

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

Offset: 1

N. J. A. Sloane, Jan 21 2020

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 iSo this is a permutation of the positive integers.
From Rémy Sigrist, Jan 23 2020: (Start)
The sequence can naturally be seen as an irregular table where:
- the n-th row has A331442(n) = 1 + A331619(T(n, 1)) terms, and
- T(n, k+1) = 2*T(n, k) for k = 1..A331442(n)-1.
(End)

			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, ...

Eric Angelini, Posting to Math Fun Mailing List, Jan 04 2020.

Rémy Sigrist, Table of n, a(n) for n = 1..10103 (first 168 chains)
Rémy Sigrist, PARI program for A331440

The inverse permutation is A331441. The lengths of the chains are given in A331442.

Cf. A331619.

PARI
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See Links section.
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A331442 Length of n-th chain in A331440.

Original entry on oeis.org

5, 8, 10, 9, 11, 11, 11, 11, 8, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 20, 16, 41, 31, 28, 28, 31, 54, 21, 9, 36, 11, 24, 23, 32, 10, 37, 42, 12, 27, 46, 16, 40, 27, 6, 19, 45, 13, 42, 42, 37, 61, 37, 98, 93, 18, 44, 33, 19, 59, 35, 54, 166, 131

Offset: 1

N. J. A. Sloane, Jan 22 2020

			When starting chain 9, A331440(77) is set to the first free number which is 17, and the chain becomes 17, 34, 68, 136, 272, 544, 1088, 2176 with length 8.

Lars Blomberg, Table of n, a(n) for n = 1..3000
Rémy Sigrist, PARI program for A331442

Cf. A331440, A331441.

PARI
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See Links section.
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Corrected a(9) and more terms from Lars Blomberg, Jan 22 2020

Showing 1-2 of 2 results.

cp's OEIS Frontend

A331440 Let S = smallest missing positive number, adjoin S, 2S, 4S, 8S, 16S, ... 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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PARI

A331442 Length of n-th chain in A331440.

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Author

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PARI

Extensions

cp's OEIS Frontend

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.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

PARI

A331442 Length of n-th chain in A331440.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

PARI

Extensions

A331440 Let S = smallest missing positive number, adjoin S, 2S, 4S, 8S, 16S, ... to the sequence until reaching a term that has S as a substring; reset S to the smallest missing positive number, repeat.