# An interesting convergence sequence generator – Kappa Function

In the article Finding Intermediate Values in Arithmetic Mean a linear formula was obtained to solve intermediate values in an Arithmetic Mean sequence. The same problem of finding the intermediate values on the Geometric Mean sequence was placed, refer to paper. During the simplification process the following sequence generator was observed.

$$\kappa(\lambda,\ n)\ =\ 2^\lambda\left(\frac{J_n}{2^n-1}\right)$$

Running the sequence for lambda, $\lambda$, values ranging from 1 to 8, and keeping $1 \le n \le \lambda$, the following sequence will be generated.

$$\begin{matrix} λ & & n = 1 & n = 2 & n = 3 & n = 4 & n = 5 & n = 6 & n = 7 & n = 8 \\ 01 & : & 2.0 \\ 02 & : & 4.0 & 2.0 \\ 03 & : & 8.0 & 4.0 & 6.0\\ 04 & : & 16.0 & 8.0 & 12.0 & 10.0\\ 05 & : & 32.0 & 16.0 & 24.0 & 20.0 & 22.0\\ 06 & : & 64.0 & 32.0 & 48.0 & 40.0 & 44.0 & 42.0\\ 07 & : & 128.0 & 64.0 & 96.0 & 80.0 & 88.0 & 84.0 & 86.0\\ 08 & : & 256.0 & 128.0 & 192.0 & 160.0 & 176.0 & 168.0 & 172.0 & 170.0\\ \end{matrix}$$

When observing the sequence generated the following properties can be observed:

1. When $n = 1, \kappa = 2^\lambda$
2. When $n = \lambda, \kappa = 2J_\lambda$
3. For $n > 1$, $n$ can be found using the equation $2^{λ−(n−1)}$
4. Given that the sequence for lambda is known, it is easy to compute any value in $\lambda + 1$ using a linear evaluation

For the finite range $1 \le n \le \lambda$ the properties of the generator are already interesting. When $n$ is not within the range, the properties get more interesting.

1. For $n = 0, \kappa$ will always return 0
2. When $n -> -\infty, \kappa$ approaches negative infinity
3. When $n -> \infty, \kappa = \frac{2^{(λ+1)}}{3}$

The proofs for the listed properties can be found in the paper, Optimisation to Geometric Mean Sequence Special Case Solver

An implementation of the Kappa function is available in github.