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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.

%I A382632 #9 Apr 06 2025 23:06:36
%S A382632 9,90,125,153,189,440,819,989,1295,1394,1484,1701,2079,2448,2925,3024,
%T A382632 4004,5453,6174,7865,8910,13509,13689,13923,16235,19683,20294,21824,
%U A382632 24804,26649,32760,33488,37169,37925,39024,40733,42704,44225,44289,47915,48734,52325,97335,101870
%N 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.
%C A382632 For all known terms (n<157), there is only one solution, except for 125 and 158949 which both have 2 solutions.
%C A382632 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.
%H A382632 Daniel Mondot, <a href="/A382632/b382632.txt">Table of n, a(n) for n = 1..156</a>
%H A382632 Allan Gottlieb, <a href="https://cs.nyu.edu/~gottlieb/tr/back-issues/2000s/2003/5-dec.pdf">Puzzle Corner</a>, Technology Review, December 2, 2003.
%e A382632 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.
%e A382632 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.
%e A382632 The 2 solutions for 125 are:
%e A382632     [3+4+...+71+72], [73+74+...+101+102] and [103+104+...+124+125 +1+2]
%e A382632 and [58+...+92], [93+...+117] and [118+...+125 + 1+...+58], all sides 2625 in length.
%Y A382632 Cf A380867, A380868, A382268, A382605.
%K A382632 nonn
%O A382632 1,1
%A A382632 _Daniel Mondot_, Apr 01 2025