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For the homogeneous linear recurrence relation, substitute $y_k=r^k$. Its characteristic equation of a linear recurrence is $r^2-r+1=0$, with roots $e^{i\pi /3}$ and $e^{-i\pi /3}$. Taking real linear combinations givesThese two real sequences are linearly independent: their values at $k=1,2$ form a matrix with determinant $\sqrt{3}/2\ne 0$. Thus their two constants account uniquely for arbitrary first two values.

Let $P_k={\sum }_{n=1}^k{a}_n/(k-n+1)$ and reindex as $P_k={\sum }_{j=1}^k{a}_{k-j+1}/j$. In $P_{k+2}-P_{k+1}+P_k$, each term with $1\le j\le k$ has coefficientThe term $j=k+1$ is $(a_2-a_1)/(k+1)=0$, and the remaining term is $a_1/(k+2)=1/(k+2)$. This explicitly proves that $P_k$ is the required particular solution, a discrete Green convolution for the period-six recurrence.

The unit-response coefficients have the homogeneous form just found. The values $a_1=a_2=1$ give zero cosine coefficient and sine coefficient $2/\sqrt{3}$, soSubtracting this particular solution from any other solution leaves a homogeneous recurrence. Thus the complete answer is