A343760 Numbers whose digits can be the lengths of the sides of a polygon.

111, 122, 133, 144, 155, 166, 177, 188, 199, 212, 221, 222, 223, 232, 233, 234, 243, 244, 245, 254, 255, 256, 265, 266, 267, 276, 277, 278, 287, 288, 289, 298, 299, 313, 322, 323, 324, 331, 332, 333, 334, 335, 342, 343, 344

Offset: 1

Author

John R Phelan, Apr 27 2021

Keywords

• base
• nonn

Comments

The length of each side must be greater than 0 and less than the sum of the other sides.

Subset of the 0-free numbers, A052382, and eventually contains all the 0-free numbers > 9111111111.

Examples

			110 is not a term since the 3rd side has a length of 0.
111 is a term since a polygon (in this case a triangle) can have sides of length 1,1,1.
112 is not a term since the length of the 3rd side is not less than the sum of the other two sides.
		

Links

• Maths StackExchange, The no. of possible 3 digit number abc if a,b,c are the sides of a isosceles triangle.
• Wikipedia, Integer Triangle

Crossrefs

Cf. A052382.

Programs

Java

public class A343760 {
   public static void main(String[] args) {
      for (long n = 1; n < 1000; n++) {
         if (is(n)) {
            System.out.print(n + ", ");
         }
      }
   }
   public static boolean is(long n) {
      String s = String.valueOf(n);
      if (n < 0 || s.contains("0")) {
         return false;
      }
      int perimeter = 0;
      char[] sides = s.toCharArray();
      for (int i = 0; i < sides.length; i++) {
         sides[i] -= '0';
         perimeter += sides[i];
      }
      for (int side : sides) {
         if (perimeter <= 2 * side) {
            return false;
         }
      }
      return true;
   }
}

Mathematica

Select[Range[111, 344], AllTrue[TakeDrop[#, 1] & /@ Permutations@ IntegerDigits[#], First[#1] < Total[#2] & @@ # &] &] (* Michael De Vlieger, May 01 2021 *)