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Take the decimal expansion of n, say n = d_1 d_2 ... d_k. We can choose to map it to any number that can be obtained by the following process. Take any substring d_i, d_{i+1}..., d_j that does not begin with 0. If the number represented by this substring is odd, replace it with twice the number. If it is even either halve it or double it.

The substring may increase or decrease in length. We do not pad it with zeros if it decreases in length.

For example, if n = 20129, then by acting on single-digit substrings we get 10129, 40129, 20229, 20119, 20149, 201218. Acting on 2-digit substrings we get in addition 2069 (halve the 12!), 20249, 20158. From 3-digit substrings we also get 40229, 20258; from 4-digit substrings we get 40249; and from 5-digit substrings we get 40258.

Eric Angelini asks what is the smallest number of steps needed to reach n if we start at 1 and repeatedly apply this process? We can reach 2 in 1 step, 4 in 2 steps, 13 in five steps, and so on.

Lars Blomberg has shown, by considering just the final digit of the numbers in the trajectory, that no number ending in 0 or 5 can be reached from 1. All other numbers can be reached (cf. A323454) - see proof below.

Update, Jan 15 2019: Lorenzo Angelini has found that 3 can be reached from 1 in 11 steps: 1, 2, 4, 8, 16, 112, 56, 28, 14, 12, 6, 3. No shorter path is possible.

From N. J. A. Sloane, Jan 16 2019: (Start)

Theorem: If k > 1 does not end in 0 or 5 then it can be reached from 1.

Proof: Suppose not, and let k be the smallest such number. Note that the allowed operations are invertible: if a -> b then also b -> a. So that means that

*** all the descendants of k must be bigger than k ***

(if there was a descendant < k, then it would also be unreachable from 1, which is a contradiction to k being the smallest).

All digits of k must be odd (if there were an even digit > 0, halve it and get a smaller number; if there is a zero digit, say we see a0, then we halve a0 and get a smaller number).

If all the digits of k are 1, do 111...1 -> 111...2 -> 55..56, a smaller number.

If there is a digit 3, 7, or 9, we know we can get that single digit down to 1 (see A323454), again a contradiction.

But all the digits can't be 5. QED (End)

The differs from the first version (in A323286) in that now n can be reached from n (by using the empty string).

This slight modification of the definition makes the analysis simpler.

The number of numbers that can be reached from n in one step is A323287(n)+1.

The minimal number of steps to reach n starting at 1 is still given by A323454.

This is equally the minimal number of steps to reach n from 1 using "Choix de Bruxelles", version 1 (cf. A323286), or -1 if n cannot be reached.

n cannot be reached if its final digit is 0 or 5, but all other numbers can be reached (see comments in A323286).

Equally, this is the total number of distinct numbers that can be obtained by starting with 1 and applying the "Choix de Bruxelles", version 1 (A323286) operation at most n times.

Also, largest number that can be obtained by starting with 1 and applying the original "Choix de Bruxelles" version 1 operation (as defined in A323286) at most n times.

a(n) is the largest number that can be obtained by applying Choix de Bruxelles (version 2) to all the numbers that can be reached from 1 by applying it n-1 times.

a(n+1) >= A323460(a(n)) (but equality does not always hold). See A307635. - Rémy Sigrist, Jan 15 2019

This is the number of terms in row n of the irregular triangle in A323286.

This is one less than the number of different numbers that can be obtained from (the decimal expansion of) n by one step of the Choix de Bruxelles, version 2 (A323460) operation. In other words, this is one less than the number of terms in row n of the irregular triangle in A323460.

Equally, this is the largest number that can be obtained from the "Choix de Bruxelles", version 1 (A323286) operation applied to n.

Maximal element in row n of irregular triangle in A323460 (or, equally, A323286).

Conjecture: If n contains no digit >= 5, then a(n) = 2*n; otherwise, a(n) is obtained from n by doubling the substring from the last digit >= 5 to the last digit. - Charlie Neder, Jan 19 2019. (This is true. - N. J. A. Sloane, Jan 22 2019)

Corollary: a(n)/n < 10 for all n, and a(n) = 10 - 1/k + O(1/k^2) for n = 10*k+5. - N. J. A. Sloane, Jan 23 2019

The high-water marks for a(n)/n occur at n = 1,15,25,35,45,..., cf. A017329. - N. J. A. Sloane, Jan 23 2019

This is equally the minimal number of steps to reach 5*n from 5 using "Choix de Bruxelles" version 1 (cf. A323286). - N. J. A. Sloane, Jan 24 2019

The sequence is well defined as we can reach any multiple of 5 starting from 5.

The Choix de Bruxelles doubles or halves some decimal digit substring and rows of A323286 are all ways this can be done.

So a(n) is the smallest term of the row a(n-1) of A323286 which is not among {a(0..n-1)}.

The sequence is finite since having reached 18 -> 9 the sole Choix for 9 would be back to 18, which is already in the sequence.

At a given term t, the Choix de Bruxelles with factor 13 can choose to multiply any decimal digit substring (not starting 0) of t by 13, or divide by 13 if that substring is divisible by 13.

These choices on substrings give various possible next values and here take the smallest not yet in the sequence.

The sequence is finite and ends at a(6851) = 7, since the sole next Choix there is multiplication by 13 to 91, but 91 is already in the sequence at the preceding a(6850) = 91.