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

3 Answers
Jul 30, 2015

2727

Explanation:

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 omega = -1/2+i sqrt(3)/2ω=12+i32

This has the property that omega^3 = 1ω3=1

Then we can write:

a_0 = 1a0=1

a_(i+1) = ((omega^i - omega)(omega^i - omega^2))/((1-omega)(1-omega^2))a_i^2+((omega^i-omega^2)(omega^i-1))/((omega-omega^2)(omega-1))(a_i+2)+((omega^i-1)(omega^i-omega))/((omega^2-1)(omega^2-omega))(a_i-1)ai+1=(ωiω)(ωiω2)(1ω)(1ω2)a2i+(ωiω2)(ωi1)(ωω2)(ω1)(ai+2)+(ωi1)(ωiω)(ω21)(ω2ω)(ai1)

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

When i = 0i=0 modulo 33, then:

((omega^i - omega)(omega^i - omega^2))/((1-omega)(1-omega^2)) = ((1 - omega)(1 - omega^2))/((1-omega)(1-omega^2)) = 1(ωiω)(ωiω2)(1ω)(1ω2)=(1ω)(1ω2)(1ω)(1ω2)=1

((omega^i-omega^2)(omega^i-1))/((omega-omega^2)(omega-1)) = ((1-omega^2)(1-1))/((1-omega^2)(omega-1)) = 0(ωiω2)(ωi1)(ωω2)(ω1)=(1ω2)(11)(1ω2)(ω1)=0

((omega^i-1)(omega^i-omega))/((omega^2-1)(omega^2-omega)) = ((1-1)(1-omega))/((omega^2-1)(omega^2-omega)) =0(ωi1)(ωiω)(ω21)(ω2ω)=(11)(1ω)(ω21)(ω2ω)=0

When i = 1i=1 modulo 33, then these coefficient expressions work out as 00, 11 and 00.

When i=2i=2 modulo 33, then these coefficient expressions work out as 00, 00 and 11.

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

Jul 31, 2015

anything you want it to be

Explanation:

Since the given sequence is obviously not arithmetic, geometric, nor exponential
one (?obvious?) solution would be to generate a polynomial function with these values as the function values for their position in the sequence:
color(white)("XXXX")XXXXf(1)=1f(1)=1
color(white)("XXXX")XXXXf(2)=1f(2)=1
color(white)("XXXX")XXXXf(3)=3f(3)=3
color(white)("XXXX")XXXXf(4)=2f(4)=2
and so on.

Given a set of nn data points (with no duplication of the xx value) it is always possible to generate a polynomial of degree nn or greater that passes through each of the points.

Jul 31, 2015

151?151?

Explanation:

As Alan says, you can construct a polynomial of degree nn for any sequence of nn values. Actually, degree n-1n1 is sufficient.

Let's look at the differences and differences of differences...

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We can use the first element of each row of differences (as enclosed by the green rectangle) to construct a polynomial for the sequence:

a_n = 1/(0!) + (0*n)/(1!) + (2*n(n-1))/(2!)-(5*n(n-1)(n-2))/(3!)+(11*n(n-1)(n-2)(n-3))/(4!)-(20n(n-1)(n-2)(n-3)(n-4))/(5!)+(29*n(n-1)(n-2)(n-3)(n-4)(n-5))/(6!)-(11n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6))/(7!)

If a sequence was really constructed using a polynomial then you would expect to encounter a row of 0's somewhat earlier, giving a simpler polynomial.

In our case that does not happen, but see what happens when we add an extra row and reconstruct the next term of the sequence. It's somewhat easier than substituting 8 into the polynomial.

So for the polynomial of lowest degree that generates the sequence as given, the next term would be 151. But as Alan says, you could choose any value and make a polynomial to fit.