OFFSET
1,1
COMMENTS
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)
REFERENCES
Eric Angelini, Email to N. J. A. Sloane, Jan 14 2019.
LINKS
Rémy Sigrist, Rows n = 1..2000, flattened
Eric Angelini, Lars Blomberg, Charlie Neder, Remy Sigrist, and N. J. A. Sloane, "Choix de Bruxelles": A New Operation on Positive Integers, arXiv:1902.01444 [math.NT], Feb 2019; Fib. Quart. 57:3 (2019), 195-200.
Brady Haran and N. J. A. Sloane, The Brussels Choice, Numberphile video (2020)
Rémy Sigrist, PARI program for A323286
EXAMPLE
The triangle begins:
2;
1, 4;
6;
2, 8;
10;
3, 12;
14;
4, 16;
18;
5, 20;
12, 21, 22;
6, 11, 14, 22, 24;
16, 23, 26;
7, 12, 18, 24, 28;
25, 30, 110;
8, 13, 26, 32, 112;
27, 34, 114;
9, 14, 28, 36, 116;
29, 38, 118;
10, 40;
11, 22, 41, 42;
11, 12, 21, 24, 42, 44;
...
PROG
(PARI) See Sigrist link.
(Python)
def cdb(n):
s, out = str(n), set()
for l in range(1, len(s)+1):
for i in range(len(s)+1-l):
if s[i] == '0': continue
t = int(s[i:i+l])
out.add(int(s[:i] + str(2*t) + s[i+l:]))
if t&1 == 0: out.add(int(s[:i] + str(t//2) + s[i+l:]))
return sorted(out)
print([c for n in range(1, 25) for c in cdb(n)]) # Michael S. Branicky, Jul 24 2022
CROSSREFS
KEYWORD
AUTHOR
N. J. A. Sloane, Jan 14 2019
EXTENSIONS
Data corrected by Rémy Sigrist, Jan 15 2019
STATUS
approved