Inspired by Thaddeus

Let's take the Generating Functions approach.

Since we only want to use odd numbers in the composition, we should look at

$f(x) = x + x^3 + x^5 + \ldots = \frac{x}{1 - x^2 }$

Since we can use an arbitrary number of partitions, we should look at

$g(x) = f(x) + f(x)^2 + \ldots = \frac{ f(x) } { 1 - f(x) }$

This is equal to $g(x) = \frac{ x } { 1 - x - x^2 }$ . We recognize this as the generating functions for the Fibonacci numbers, hence the coefficient of $x^{10}$ in $g(x)$ is $F_{10} = 55$ .

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Now that you know it's a pattern of Fibonacci numbers, can you prove it in another way?

Let $P(n)$ be odd product reductions of composition of n. Depending on the first number inserted $P(n)$ can be written as,

$P(2k) = \sum_{i=1}^{k} P(2i-1)$ ,

$P(2k+1) = 1 + \sum_{i=1}^{k} P(2i)$ .

$P(2k+1) = P(2k) + P(2k-1)$ (As, $P(2k-1) = 1 + \sum_{i=1}^{k-1} P(i)$ ),

$P(2k) = P(2k-1) + P(2k-2)$ (As, $P(2k-2) = \sum_{i=1}^{k-1} P(i)$ ).

Hence, $P(n) = P(n-1) + P(n-2)$ , with $P(1) = 1, P(2) = 1$ .

Hence, $P(n) = F_n$ , for $n>0$ ( $F_n$ is $n^{th}$ fibonacci number. )