A382632 Numbers k such that one can make an equilateral triangle from a chain of linked rods of length 1, 2, 3, ..., k, with perimeter equal to the total length.

9, 90, 125, 153, 189, 440, 819, 989, 1295, 1394, 1484, 1701, 2079, 2448, 2925, 3024, 4004, 5453, 6174, 7865, 8910, 13509, 13689, 13923, 16235, 19683, 20294, 21824, 24804, 26649, 32760, 33488, 37169, 37925, 39024, 40733, 42704, 44225, 44289, 47915, 48734, 52325, 97335, 101870

Offset: 1

Author

Daniel Mondot, Apr 01 2025

Keywords

• nonn

Comments

For all known terms (n<157), there is only one solution, except for 125 and 158949 which both have 2 solutions.

Conjecture: Out of some linked rods of length 1, 2...k, we can fold them in half (digon), or make equilateral triangles, but no other regular polygons (squares, regular pentagons, etc...) can be made.

Examples

			For k=9, the sides of the triangle are 15 in length: [4+5+6], [7+8] and [9+1+2+3]. Therefore 9 is in the sequence.
For k=90, the sides of the triangle are 1365 in length: [16+...+54], [55+...+75] and [76+...+90 + 1+...+15]. Therefore 90 is in the sequence.
The 2 solutions for 125 are:
    [3+4+...+71+72], [73+74+...+101+102] and [103+104+...+124+125 +1+2]
and [58+...+92], [93+...+117] and [118+...+125 + 1+...+58], all sides 2625 in length.
		

Links

• Daniel Mondot, Table of n, a(n) for n = 1..156
• Allan Gottlieb, Puzzle Corner, Technology Review, December 2, 2003.

Crossrefs

Cf A380867, A380868, A382268, A382605.