For my research interest, I came accross with OEIS sequence A280194
In particular I am interested how is obtained the value $36203$ in the sequence
Expansion of:
$\frac{1}{1-\sum_{k>=1}{\mu(k)^2\cdot x^k}}$ where $\mu$ is the Moebius function.
Let $a(n)$ denote a term in the sequence OEIS A280194. How is calculated for example $a(4)=7$?
You asked
The OEIS sequence A280194 entry states:
This comes from the comment
and the example shows the $7$ compositions of $4$ into squarefree parts in reverse lexicographic order. To show that $a(5)=14$ list all of the 14 compositions of $5$ into squarefree parts
$$ \texttt{ 5 = 3+2 = 3+1+1 = 2+3 = 2+2+1 = 2+1+2 = 2+1+1+1 = 1+3+1 =}\\ \texttt{ 1+2+2 = 1+2+1+1 = 1+1+3 = 1+1+2+1 = 1+1+1+2 = 1+1+1+1+1}. $$
The interpretation as compositions into squarefree parts comes almost directly from the name of the sequence
The square of the Moebius function is the indicator of squarefree numbers.
A simple example is the case of $\,1/(1 - x - x^2)\,$ which is the generating function of the Fibonacci numbers with $\,a(n)=0\,$ if $\,n<0,\,$ $\,a(0) = 1,\,$ and $\,a(n) = a(n-1)+a(n-2) \,$ if $\,n>0.\,$ The interpretation is that $\,a(n)\,$ is the number of compositions of $n$ into parts of $1$ and $2$ only.
Your question
asks how to evaluate $a(17) = 36203.$ In this case it is possible to write a program to systematically list all $36203$ compositions of $17$. An easier method is to use the recursion which comes from the generating function similar to the Fibonacci example. That is
$$ \texttt{ a(n) = a(n-1) + a(n-2) + a(n-3) + a(n-5) + a(n-6) + a(n-7) + a(n-10)}\\ \texttt {+ a(n-11) + a(n-13) + a(n-14) + a(n-15) + a(n-17) +}\cdots. $$
The easiest method is to use a Computer Algebra System to do the calculation with power series.