For example for [6], $d(1) = \gcd\{3, 5, 6,\, ...\} = 1$.
What do $3,5,6$ calculated from?
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$\begingroup$ Seems to be times $k$ when $P_k(1,1)>0$ where $P_k(i,j) = \mathsf P(X_k = j|X_0 = i)$. $\endgroup$– SBFJan 3, 2013 at 12:08
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1 Answer
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For other readers' reference, this question refers to the Markov chain in [6] shown here:
and $d(i)$ is the $\gcd$ of those $n>0$ for which there is a path from $i$ to $i$ of length $n$.
How can you get from $1$ to $1$ in $>0$ steps?
Well you can go $1 \to 2 \to 3 \to 1$. That's $3$ steps.
Or you can go $1 \to 2 \to 3 \to 4 \to 3 \to 1$. That's $5$ steps.
Or you can go $1 \to 2 \to 3 \to 1 \to 2 \to 3 \to 1$. That's $6$ steps.
And so on.
answered Jan 3, 2013 at 12:09