What is the next number in this sequence 1, 1, 3, 2, 4, 6, 5, 25?

Question

Yareli Hendrix · Accepted Answer

This is not a very mathematically significant kind of sequence.

The steps are:

1) Square the previous term.
2) Add 2 to the previous term.
3) Subtract 1 from the previous term.
(repeat)

Is there a single algebraic formula to describe this iterative process?

Consider

ω = - \frac{1}{2} + i \frac{\sqrt{3}}{2}

This has the property that

ω^{3} = 1

Then we can write:

a_{0} = 1

a_{i + 1} = \frac{(ω^{i} - ω) (ω^{i} - ω^{2})}{(1 - ω) (1 - ω^{2})} a_{i}^{2} + \frac{(ω^{i} - ω^{2}) (ω^{i} - 1)}{(ω - ω^{2}) (ω - 1)} (a_{i} + 2) + \frac{(ω^{i} - 1) (ω^{i} - ω)}{(ω^{2} - 1) (ω^{2} - ω)} (a_{i} - 1)

This can be simplified, but it helps to have it in this formulation so you can understand how it works.

When i=0 modulo 3, then:

\frac{(ω^{i} - ω) (ω^{i} - ω^{2})}{(1 - ω) (1 - ω^{2})} = \frac{(1 - ω) (1 - ω^{2})}{(1 - ω) (1 - ω^{2})} = 1

\frac{(ω^{i} - ω^{2}) (ω^{i} - 1)}{(ω - ω^{2}) (ω - 1)} = \frac{(1 - ω^{2}) (1 - 1)}{(1 - ω^{2}) (ω - 1)} = 0

\frac{(ω^{i} - 1) (ω^{i} - ω)}{(ω^{2} - 1) (ω^{2} - ω)} = \frac{(1 - 1) (1 - ω)}{(ω^{2} - 1) (ω^{2} - ω)} = 0

When i=1 modulo 3, then these coefficient expressions work out as 0, 1 and 0.

When i=2 modulo 3, then these coefficient expressions work out as 0, 0 and 1.

So we use these to pick out each of the three rules cyclically.

What is the next number in this sequence 1, 1, 3, 2, 4, 6, 5, 25?

Answered question

Answer & Explanation

New Questions in Algebra I