I think the claim in this form is wrong. The eigenvalues are not integral. For example, with $n=4$ the matrix is
$$
A=\begin{pmatrix}
0 & 4 & 0 & 0\cr
1 & 0 & 3 & 0\cr
0 & 2 & 0 & 2 \cr
0 & 0 & 3 & 0
\end{pmatrix}.
$$

The characteristic polynomial of this matrix is 
$\chi (t)=t^4-16 t^2 +24$, which has no integral roots.
Am I overlooking something ?

Edit: just visited the site http://math.nist.gov/MatrixMarket/deli/Clement/
Here I saw that the upper diagonal must be $3,2,1$, not $4,3,2$. Then everything is OK.
So $y_k=n-k$ rather than $y_k=n-k-1$, what you wrote. We have a recursion for $\chi(t)=\det(t\cdot id-A)$, see http://mathoverflow.net/questions/131527/eigenvalues-of-symmetric-tridiagonal-matrices, from which the result can be derived easily.