Numbers Partitioned into Six Hexagonal Numbers

Theorem

The integers $11$ and $26$ cannot be represented by the sum of less than $6$ hexagonal numbers.

Proof

Recall the sequence of hexagonal numbers:

The sequence of hexagonal numbers, for $n \in \Z_{\ge 0}$, begins:

$0, 1, 6, 15, 28, 45, 66, 91, 120, 153, 190, 231, \ldots$

Hence:

\(\ds 11\)

\(=\)

\(\ds 6 + 1 + 1 + 1 + 1 + 1\)

\(\ds 26\)

\(=\)

\(\ds 6 + 6 + 6 + 6 + 1 + 1\)

$\blacksquare$

Sources

• 1994: Richard K. Guy: Unsolved Problems in Number Theory (2nd ed.)
• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $26$