What is the smallest $k$ such that every $n \in \mathbb{N}$ can be represented by a sum of exactly $k$ triangular numbers? For the sake of simplicity, I will assume $0$ is a triangular number.
I've been able to establish that $3 \le k \le 8$. The lower bound was found through experimentation. The upper bound is a consequence of Waring's problem, and also the neat property that two consecutive triangular numbers always sum to a square.
Any ideas on how to obtain more reasonable bounds on $k$? Or perhaps even some obvious method to determine $k$ which I have overlooked?
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Best Answer
THREE.
Multiply by 8. Add 3. It is a theorem of Gauss and Legendre that every $8n+3$ is the sum of three odd squares.
SEE, from the book The Sensual Quadratic Form by John Horton Conway, pages 138,139,140:
A list from a book by Dickson, giving sums $f(x,y,z) = a x^2 + b y^2 + c z^2$ with $1 \leq a \leq b \leq c,$ and all the numbers that cannot be expressed by that one. So, $a=b=c=1,$ we find that the sum of three squares $x^2 + y^2 + z^2$ represents every positive integer not of the form $4^k \, (8n+7).$ In particular, all $8n + 3$ are represented.