Sum of three bounded triangular numbers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T22:37:20Z http://mathoverflow.net/feeds/question/108075 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108075/sum-of-three-bounded-triangular-numbers Sum of three bounded triangular numbers Aaron Meyerowitz 2012-09-25T16:45:27Z 2012-10-03T04:57:34Z <p>Every nonnegative integer can be written (eventually in many ways) as a sum of three triangular numbers by the <a href="http://www3.ntu.edu.sg/home/rsinai/Site/Publications_files/On%20the%20representation%20of%20integers.pdf" rel="nofollow">Gauss Eureka theorem</a>.</p> <blockquote> <p>What is the smallest positive integer $n=n_m$ which can <strong>not</strong> be written in the form $$n=\binom{a}{2}+\binom{b}{2}+\binom{c}{2}.$$ subject to $\max(a,b,c) \le m?$</p> </blockquote> <p>The answers for $m$ from $1$ to 60 are</p> <p>$\begin {array}{cccccccccc} 1&amp;4&amp;8&amp;11&amp;24&amp;29&amp;29&amp;47&amp;68&amp;68 \\95&amp;99&amp;137&amp;141&amp;173&amp;173&amp;245&amp;281&amp;314&amp;314 \\314&amp;407&amp;419&amp;419&amp;470&amp;470&amp;617&amp;617&amp;711&amp;800 \\863&amp;911&amp;911&amp;911&amp;911&amp;1118&amp;1118&amp;1118&amp;1118&amp;1118 \\1383&amp;1433&amp;1433&amp;1679&amp;1679&amp;1679&amp;1868&amp;1868&amp;1868&amp;1868 \\1868&amp;2360&amp;2493&amp;2493&amp;2519&amp;2925&amp;3044&amp;3044&amp;3098&amp;3098 \end {array}$</p> <p>The sequence does not seem to be in the OEIS even with a superseeker search. It just seems a curious sequence. Any information would be welcome. What bounds or asymptotics can be established? It would appear that $n_m \lt m^2$ although $m_{105}=11018 \lt 11025=105^2$ leaves some doubt. I'd conjecture (rashly) that $$\limsup \frac{n_m}{m^2}=1$$ but $n_{110}=n_{111}=n_{112}=n_{113}=11625$ and $\frac{11625}{113^3} \approx 0.91$ so the $\liminf$ might be less. I wonder what explains the repeated values and what can be said about them.</p> <p><strong>UPDATE</strong> Noam makes a nice argument that $\lim \frac{n_m}{m^2}=3/2.$ I'd accept it if it was an answer and not a comment. Let me spell out that $8\binom{t}{2}+1=(2t+1)^2$ so $n=\binom{a}{2}+\binom{b}{2}+\binom{c}{2}$ exactly if $8n+3=(2a+1)^2+(2b+1)^2+(2c+1)^2$ and this allows one to pull in results on sums of squares.</p> <p>I still don't see why things such as $n_{36}=n_{37}=n_{38}=n_{39}=n_{40}$ happen, but I have not thought about it very deeply. </p> <p>For any fixed $k$, $n_m \lt \frac{3m^2}{2}-km$ with finitely many exceptions. so one could wonder about things like $\frac{3m^2}{2}-\sqrt{m^3}.$ However this result was great for <a href="http://mathoverflow.net/questions/107944" rel="nofollow">the problem</a> I wanted to apply it to. I would have mentioned the connection sooner but the answer came before I ggot to that.</p> http://mathoverflow.net/questions/108075/sum-of-three-bounded-triangular-numbers/108086#108086 Answer by Robert Israel for Sum of three bounded triangular numbers Robert Israel 2012-09-25T18:30:29Z 2012-09-25T18:30:29Z <p>I get $n_{m} > m^2$ for the following values of $m$ (up to 322): \eqalign{ &amp;118, 139, 140, 141, 152, 153, 176, 177, 179, 180, 182, 183, 184, 185, 186, 188,\cr &amp;189, 190, 191, 192, 193, 194, 196, 197, 198, 199, 200, 201, 202, 203, 209, 210,\cr &amp;220, 221, 222, 223, 224, 225, 227, 228, 229, 230, 231, 232, 233, 234, 235, 236,\cr &amp;242, 243, 244, 249, 250, 251, 252, 253, 254, 259, 260, 261, 262, 263, 264, 265,\cr &amp;266, 267, 270, 271, 272, 273, 274, 275, 278, 279, 280, 281, 282, 283, 284, 285,\cr &amp;286, 287, 288, 289, 290, 291, 292, 293, 294, 295, 296, 297, 298, 299, 300, 301,\cr &amp;302, 303 }</p>