

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^(ij)*S, so T was not a missing number. If i<j then S = 2^(ji)*T, and T would have been a smaller choice for S. QED
So 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 nth 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)


REFERENCES

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


LINKS

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


EXAMPLE

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, ...


PROG

(PARI) See Links section.


CROSSREFS

The inverse permutation is A331441. The lengths of the chains are given in A331442.
Cf. A331619.
Sequence in context: A036122 A050124 A101943 * A247243 A341819 A352387
Adjacent sequences: A331437 A331438 A331439 * A331441 A331442 A331443


KEYWORD

nonn,base,tabf


AUTHOR

N. J. A. Sloane, Jan 21 2020


STATUS

approved



