Sum of three bounded triangular numbers - MathOverflow most recent 30 from http://mathoverflow.net2013-05-24T22:37:20Zhttp://mathoverflow.net/feeds/question/108075http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/108075/sum-of-three-bounded-triangular-numbersSum of three bounded triangular numbersAaron Meyerowitz2012-09-25T16:45:27Z2012-10-03T04:57:34Z
<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&4&8&11&24&29&29&47&68&68
\\95&99&137&141&173&173&245&281&314&314
\\314&407&419&419&470&470&617&617&711&800
\\863&911&911&911&911&1118&1118&1118&1118&1118
\\1383&1433&1433&1679&1679&1679&1868&1868&1868&1868
\\1868&2360&2493&2493&2519&2925&3044&3044&3098&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#108086Answer by Robert Israel for Sum of three bounded triangular numbersRobert Israel2012-09-25T18:30:29Z2012-09-25T18:30:29Z<p>I get $n_{m} > m^2$ for the following values of $m$ (up to 322):
$$\eqalign{ &118, 139, 140, 141, 152, 153, 176, 177, 179, 180, 182, 183, 184, 185, 186, 188,\cr &189, 190, 191, 192, 193, 194, 196, 197, 198, 199, 200, 201, 202, 203, 209,
210,\cr &220, 221, 222, 223, 224, 225, 227, 228, 229, 230, 231, 232, 233, 234, 235,
236,\cr &242, 243, 244, 249, 250, 251, 252, 253, 254, 259, 260, 261, 262, 263, 264,
265,\cr &266, 267, 270, 271, 272, 273, 274, 275, 278, 279, 280, 281, 282, 283, 284,
285,\cr &286, 287, 288, 289, 290, 291, 292, 293, 294, 295, 296, 297, 298, 299, 300,
301,\cr &302, 303 }$$</p>