The product of four consecutive Fibonacci numbers equals the product of two consecutive integers.
For example,
3 × 5 × 8 × 13 = 39 × 40.
I ran across this theorem in a note [1] that says “The product of any four consecutive Fibonacci numbers is twice a triangular number.” Since triangular numbers have the form n(n + 1)/2, twice a triangular number is the product of two consecutive integers.
The note also gives a way to find the numbers on the right hand side. We have
Fn Fn+1 Fn+2 Fn+3 = m(m + 1)
where m equals
Fn+1 Fn+2
if n is odd and
Fn Fn+3
if n is even.
In the example at the top, 3 is the 4th Fibonacci number, so n = 4. Since 4 is even, m is the product of the 4th and 7th Fibonacci numbers, i.e. m = 3 × 13 = 39.
More Fibonacci posts
- Fibonacci meets Pythagoras
- Certified Fibonacci numbers
- Turning trig identities into Fibonacci identities
[1] K. B. Subramaniam. On a link between Triangular and Fibonacci numbers. The Mathematical Gazette, Vol. 103, No. 558 (November 2019), p. 489.
Very cool! Another way to get to this is analyzing the product F(n)*F(n+3)-F(n+1)*F(n+2) in terms of F(n-2) and F(n-1), which leads to a relationship from consecutive quadruplets to consecutive triplets
F(n)*F(n+3)-F(n+1)*F(n+2) = F(n-2)*F(n) – F(n-1)^2
Defining this triplet function as J(n) = F(n-1)*F(n+1) – F(n-2)^2, we can then descend the triplets:
J(n-1) = F(n-2)*F(n) – F(n-1)^2 = F(n-2)*(F(n-2)+F(n-1)) – F(n-1)^2 = F(n-2)^2 + F(n-2)F(n-1) – F(n-1)^2
= F(n-2)^2 + F(n-1) (F(n-2)- F(n-1)) = F(n-2)^2 – F(n-1) F(n-3) = – J(n-2).
And going down leads to J(2) = 2*0 – 1 = -1.