Algorithm to produce number series - algorithm

Question

I am not sure how to attack this problem... I tried many things, and it seems to be that it shouldn't be so difficult, but not getting there...
Is it possible to create a function "series ( _x )", that produces this :
The function for example should be myfunction( 11 ) => 211

Answer 1

The terms become suffix for the next terms. See below picture for more clarity. The boxes with same color gets repeated. So, we could just keep prepending 1 and 2 for previous results.
Code(In java):
public class Solution {
public static void main(String[] args) {
List<String> ans = solve(10);
for(int i=0;i<ans.size();++i) System.out.println(ans.get(i));
}
private static List<String> solve(int terms){
List<String> ans = new ArrayList<>();
String[] digits = new String[]{"1","2"};
ans.add("1");
if(terms == 1) return ans;
ans.add("2");
if(terms == 2) return ans;
List<String> final_result = new ArrayList<>();
final_result.addAll(ans);
terms -= 2;//since 2 numbers are already added
while(terms > 0){
List<String> temp = new ArrayList<>();
for(String s : digits){
for(int j=0;j<ans.size() && terms > 0;++j){
temp.add(s + ans.get(j));
terms--;
}
}
ans = temp;
final_result.addAll(ans);
}
return final_result;
}
}

Answer 2

This hint should help you... It isn't quite binary, but it is close. Let me know if you need any further help
0 -> - -> -
1 -> - -> -
10 -> 0 -> 1
11 -> 1 -> 2
100 -> 00 -> 11
101 -> 01 -> 12
110 -> 10 -> 21
111 -> 11 -> 22
1000 -> 000 -> 111
1001 -> 001 -> 112
1010 -> 010 -> 121
1011 -> 011 -> 122
1100 -> 100 -> 211
1101 -> 101 -> 212
1110 -> 110 -> 221
1111 -> 111 -> 222
Edit: I didn't like the way I ordered the columns, so I swapped 2 and 3
Python approach
First thing that we need to do is produce binary strings
in Python this can be done with bin(number)
However this will return a number in the form 0b101
We can easily strip away the 0b from the beginning though by telling python that we dont want the first two characters, but we want all the rest of them. The code for that is: bin(number)[2:] left side of the : says start two spaces in, and since the right side is blank go to the end
Now we have the binary numbers, but we need to strip away the first number. Luckily we already know how to strip away leading characters so change that line to bin(number)[3:].
All that is left to do now is add one to every position in the number. To do that lets make a new string and add each character from our other string to it after incrementing it by one.
# we already had this
binary = bin(user_in + 1)[3:]
new = ""
for char in binary:
# add to the string the character + 1
new += str(int(char) + 1)
And we are done. That snippet will convert from decimal to whatever this system is. One thing you might notice is that this solution will be offset by one (2 will be 1, 3 will be 2) we can fix this by simply adding one to user input before we begin.
final code with some convenience (a while loop and print statement)
while True:
user_in = int(input("enter number: "))
binary = bin(user_in + 1)[3:]
new = ""
for char in binary:
new += str(int(char) + 1)
print(user_in, "\t->\t", binary, "\t->\t", new)

Answer 3

According to A000055
We should perform 3 steps:
Convert value + 1 to base 2
Remove 1st 1
Add 1 to the rest digits
For instance, for 11 we have
Converting 11 + 1 == 12 to binary: 1100
Removing 1st 1: 100
Adding 1 to the rest digits: 211
So 11 has 211 representation.
C# code:
private static String MyCode(int value) =>
string.Concat(Convert
.ToString(value + 1, 2) // To Binary
.Skip(1) // Skip (Remove) 1st 1
.Select(c => (char)(c + 1))); // Add 1 to the rest digits
Demo:
var result = Enumerable
.Range(1, 22)
.Select(value => $"{MyCode(value),4} : {value,2}");
Console.Write(string.Join(Emvironment.NewLine, result));
Outcome:
1 : 1
2 : 2
11 : 3
12 : 4
21 : 5
22 : 6
111 : 7
112 : 8
121 : 9
122 : 10
211 : 11
212 : 12
221 : 13
222 : 14
1111 : 15
1112 : 16
1121 : 17
1122 : 18
1211 : 19
1212 : 20
1221 : 21
1222 : 22

Answer 4

In VB.NET, showing both the counting in base-3 and OEIS formula ways, with no attempts at optimisation:
Module Module1
Function OEIS_A007931(n As Integer) As Integer
' From https://oeis.org/A007931
Dim m = Math.Floor(Math.Log(n + 1) / Math.Log(2))
Dim x = 0
For j = 0 To m - 1
Dim b = Math.Floor((n + 1 - 2 ^ m) / (2 ^ j))
x += CInt((1 + b Mod 2) * 10 ^ j)
Next
Return x
End Function
Function ToBase3(n As Integer) As String
Dim s = ""
While n > 0
s = (n Mod 3).ToString() & s
n \= 3
End While
Return s
End Function
Function SkipZeros(n As Integer) As String
Dim i = 0
Dim num = 1
Dim s = ""
While i < n
s = ToBase3(num)
If s.IndexOf("0"c) = -1 Then
i += 1
End If
num += 1
End While
Return s
End Function
Sub Main()
Console.WriteLine("A007931 Base3 ITERATION")
For i = 1 To 22
Console.WriteLine(OEIS_A007931(i).ToString().PadLeft(7) & SkipZeros(i).PadLeft(7) & i.ToString().PadLeft(11))
Next
Console.ReadLine()
End Sub
End Module
Outputs:
A007931 Base3 ITERATION
1 1 1
2 2 2
11 11 3
12 12 4
21 21 5
22 22 6
111 111 7
112 112 8
121 121 9
122 122 10
211 211 11
212 212 12
221 221 13
222 222 14
1111 1111 15
1112 1112 16
1121 1121 17
1122 1122 18
1211 1211 19
1212 1212 20
1221 1221 21
1222 1222 22