Centered Hexagonal Number as Sum of Triangular Numbers

Theorem

Let $C_n$ be the $n$th centered hexagonal number.

Then:

$C_n = 6 T_{n - 1} + 1$

where $T_{n - 1}$ denotes the $n - 1$th triangular number.

Proof

\(\ds C_n\)

\(=\)

\(\ds 3 n \paren {n - 1} + 1\)

Closed Form for Centered Hexagonal Numbers

\(\ds \)

\(=\)

\(\ds 6 \paren {\dfrac {\paren {n - 1} n} 2} + 1\)

\(\ds \)

\(=\)

\(\ds 6 T_{n - 1} + 1\)

Closed Form for Triangular Numbers

$\blacksquare$

Visual Demonstration

Sources

• 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $37$
• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $37$