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Consider the monic orthogonal polynomials determined by the recurrence $$p_{n+1}(x)=(x-n(n+b))p_{n}(x)-n(n+a)p_{n-1}(x), \quad n\in\mathbb{N},$$ with the initial conditions $p_{-1}(x)=0$ and $p_{0}(x)=1$. The parameters $a$ and $b$ are assumed to be $a>-1$ and $b\in\mathbb{R}$. I am particularly interested in the following questions:

1) Is their measure of ortogonality expressible in terms of some special functions?

2) Is there a generating function for this family expressible in terms of some special functions?

Remark: Let me remark that I've checked the Askey scheme but found nothing. There are several polynomials in the Askey scheme whose coefficients from the three-term recurrence are polynomials in $n$ of low degree. These coefficients sometimes look similarly as above but it seems that they are never both quadratic in $n$.

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  • $\begingroup$ @ChristianRemling: There is always a positive measure wrt which the sequence of polynomials generated by the three term recurrence is orthogonal providing that the first coefficient, in this case $n(n+b)$, is real for all $n\geq0$ and the second coefficient, in this case $n(n+a)$, is positive for all $n>0$. One explanation relies on the Spectral theorem for self-adjoint operators. Here, the operator is the Jacobi operator $J$ with diagonal $n(n+b)$ and off-diagonal $\sqrt{(n+1)(n+1+a)}$, $n\geq0$. The meas. of orthogonality for the above family is closely related to the spec. meas. of $J$. $\endgroup$
    – Twi
    Commented Jan 5, 2018 at 8:29
  • $\begingroup$ Yes, you are right, the recursion can be rewritten to bring it to this form. So you want to compute the spectral measure of this Jacobi matrix, and I'd be quite surprised if that could be done explicitly (maybe in friendly special cases such as $a=0$). $\endgroup$
    – Christian Remling
    Commented Jan 5, 2018 at 20:03
  • $\begingroup$ It should be possible though to obtain some qualitative information. I suspect the spectrum will be purely discrete and bounded below (and obviously it's unbounded above). $\endgroup$
    – Christian Remling
    Commented Jan 5, 2018 at 20:28
  • $\begingroup$ @ChristianRemling: You are right. The spectrum is purely discrete. There is a criterion applicable to this case. It also seems that the operator is semi-bounded but I didn't try to prove it. I also didn't get an explicit result in any special case. $\endgroup$
    – Twi
    Commented Jan 6, 2018 at 21:37

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