Integer as Sum of Polygonal Numbers/Lemma 1

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Theorem

Let $n, m \in \N_{>0}$ such that $m \ge 3$.

Let $n < 116 m$.

Then $n$ can be expressed as a sum of at most $m + 2$ polygonal numbers of order $m + 2$.


Proof

From Closed Form for Polygonal Numbers:

$\map P {m + 2, k} = \dfrac m 2 \paren {k^2 - k} + k = m T_{k - 1} + k$

Where $T_{k - 1}$ are triangular numbers.


The first few $\paren {m + 2}$-gonal numbers less than $116 m$ are:

$0, 1, m + 2, 3 m + 3, 6 m + 4, 10 m + 5, 15 m + 6, 21 m + 7, 28 m + 8, 36 m + 9, 45 m + 10, 55 m + 11, 66 m + 12, 78 m + 13, 91 m + 14, 105 m + 15$


We show the expression of the first few numbers explicitly.

\(\ds 1\) \(=\) \(\ds 1\)
\(\ds 2\) \(=\) \(\ds 1 + 1\)
\(\ds \) \(:\) \(\ds \)
\(\ds m + 1\) \(=\) \(\ds 1 + 1 + \dots + 1\) There are $m + 1$ $1$'s
\(\ds m + 2\) \(=\) \(\ds \paren {m + 2}\)
\(\ds \) \(:\) \(\ds \)
\(\ds 2 m + 3\) \(=\) \(\ds \paren {m + 2} + 1 + 1 + \dots + 1\) There are $m + 1$ $1$'s
\(\ds 2 m + 4\) \(=\) \(\ds \paren {m + 2} + \paren {m + 2}\)
\(\ds \) \(:\) \(\ds \)
\(\ds 3 m + 2\) \(=\) \(\ds \paren {m + 2} + \paren {m + 2} + 1 + 1 + \dots + 1\) There are $m - 2$ $1$'s
\(\ds 3 m + 3\) \(=\) \(\ds \paren {3 m + 3}\)
\(\ds \) \(:\) \(\ds \)
\(\ds 4 m + 4\) \(=\) \(\ds \paren {3 m + 3} + 1 + 1 + \dots + 1\) There are $m + 1$ $1$'s
\(\ds 4 m + 5\) \(=\) \(\ds \paren {3 m + 3} + \paren {m + 2}\)
\(\ds \) \(:\) \(\ds \)
\(\ds 5 m + 5\) \(=\) \(\ds \paren {3 m + 3} + \paren {m + 2} + 1 + 1 + \dots + 1\) There are $m$ $1$'s
\(\ds 5 m + 6\) \(=\) \(\ds \paren {m + 2} + \paren {m + 2} + \paren {m + 2} + \paren {m + 2} + 1 + 1 + \dots + 1\) There are $m - 2$ $1$'s
\(\ds 5 m + 7\) \(=\) \(\ds \paren {3 m + 3} + \paren {m + 2} + \paren {m + 2}\)
\(\ds \) \(:\) \(\ds \)
\(\ds 6 m + 6\) \(=\) \(\ds \paren {3 m + 3} + \paren {m + 2} + \paren {m + 2} + 1 + 1 + \dots + 1\) There are $m - 1$ $1$'s
\(\ds 6 m + 7\) \(=\) \(\ds \paren {6 m + 4} + 1 + 1 + 1\)
\(\ds \) \(:\) \(\ds \)
\(\ds 7 m + 5\) \(=\) \(\ds \paren {6 m + 4} + 1 + 1 + \dots + 1\) There are $m + 1$ $1$'s
\(\ds 7 m + 6\) \(=\) \(\ds \paren {6 m + 4} + \paren {m + 2}\)
\(\ds \) \(:\) \(\ds \)
\(\ds 8 m + 6\) \(=\) \(\ds \paren {6 m + 4} + \paren {m + 2} + 1 + 1 + \dots + 1\) There are $m$ $1$'s
\(\ds 8 m + 7\) \(=\) \(\ds \paren {3 m + 3} + \paren {3 m + 3} + \paren {m + 2} + 1 + 1 + \dots + 1\) There are $m - 1$ $1$'s
\(\ds 8 m + 8\) \(=\) \(\ds \paren {6 m + 4} + \paren {m + 2} + \paren {m + 2}\)
\(\ds \) \(:\) \(\ds \)
\(\ds 9 m + 7\) \(=\) \(\ds \paren {6 m + 4} + \paren {m + 2} + \paren {m + 2} + 1 + 1 + \dots + 1\) There are $m - 1$ $1$'s
\(\ds 9 m + 8\) \(=\) \(\ds \paren {6 m + 4} + \paren {3 m + 3} + 1\)
\(\ds \) \(:\) \(\ds \)
\(\ds 10 m + 8\) \(=\) \(\ds \paren {6 m + 4} + \paren {3 m + 3} + 1 + 1 + \dots + 1\) There are $m$ $1$'s
\(\ds 10 m + 9\) \(=\) \(\ds \paren {10 m + 5} + 1 + 1 + 1 + 1\)
\(\ds \) \(:\) \(\ds \)
\(\ds 11 m + 10\) \(=\) \(\ds \paren {10 m + 5} + 1 + 1 + \dots + 1\) There are $m + 1$ $1$'s

and so on.

$\blacksquare$