Triangular Fibonacci Numbers
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Theorem
The only Fibonacci numbers which are also triangular are:
- $0, 1, 3, 21, 55$
This sequence is A039595 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).
Proof
\(\ds 0\) | \(=\) | \(\ds \dfrac {0 \times 1} 2\) | ||||||||||||
\(\ds 1\) | \(=\) | \(\ds \dfrac {1 \times 2} 2\) | ||||||||||||
\(\ds 3\) | \(=\) | \(\ds \dfrac {2 \times 3} 2\) | \(\ds = 1 + 2\) | |||||||||||
\(\ds 21\) | \(=\) | \(\ds \dfrac {6 \times 7} 2\) | \(\ds = 8 + 13\) | |||||||||||
\(\ds 55\) | \(=\) | \(\ds \dfrac {10 \times 11} 2\) | \(\ds = 21 + 34\) |
It remains to be shown that these are the only ones.
Let $F_n$ be the $n$th Fibonacci number.
From Odd Square is Eight Triangles Plus One, $F_n$ is triangular if and only if $8 F_n + 1$ is square.
It remains to be demonstrated that $8 F_n + 1$ is square if and only if:
- $n \in \set{\pm 1, 0, 2, 4, 8, 10}$
So, let $8 F_n + 1$ be square.
Then:
- $n \equiv \begin{cases} \pm 1 \pmod {2^5 \times 5} & : n \text { odd} \\ 0, 2, 4, 8, 10 \pmod {2^5 \times 5^2 \times 11} & : n \text { even} \end{cases}$
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Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $55$
- 1989: Luo Ming: On Triangular Fibonacci Numbers (The Fibonacci Quarterly Vol. 27, no. 2: pp. 98 – 108)
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $55$