Decomposition of $_1F_2(1+n;1,2+n;x)$

Question

I am looking for a way to decompose $_1F_2(1+n;1,2+n;z)$ for $n\in\mathbb{N}$ into either Bessel J functions or regularized confluent hypergeometric functions $_0\tilde F_1(b(n),z)$. Mathematica seems to be able to do the decomposition for $n=1,2,3,4$ but I am trying to get a closed form for any $n$. I have looked through Wolfram's functions site but have been unsuccessful in my search. This would be useful because numerical routines can deal with Bessels much faster than hypergeometrics.

Answer 1

While I've studied this hypergeometric series I've observed that

$${_1F_2}\left(\begin{array}c1+n\\1,\,2+n\end{array}\middle|\,x\right)=\sum_{k=0}^{2n}\frac{(-1)^k\,c_{n,k}\,I_{k+1 \operatorname{mod} 2}\left(2\sqrt x\right)}{x^{n-\frac k2+ \frac 12}},$$

with

$$c_{n,k}=\frac{n!(n+1)!}{\left\lfloor\frac k2 \right\rfloor!\left\lfloor\frac k2 + \frac 12 \right\rfloor!}$$

coefficients. Here $I_\alpha(x)$ denotes the modified Bessel function of the first kind.

Answer 2

I have found a somewhat simple relation based on the triangle of falling factorials, reading by rows (A068424). We have $$_1F_2(1+n;1,2+n;z)=\sum_{k=1}^{n+1}\begin{pmatrix} n+1 \\ k \end{pmatrix}k!(-1)^{k+1}\,_0\tilde F_1(;1+k;z),\,n\geq 0 \\ =\sum_{k=1}^{n+1}\begin{pmatrix} n+1 \\ k \end{pmatrix}k!(-1)^{k+1}(-z)^{-k/2}J_k(2\sqrt{-z}),\,n\geq 0.$$

Now we can use the recurrance relation to get it all in terms of 1 or 2 bessels/confluents in order to reduce computation time. By Wolfram's Functions we have $$_0\tilde F_1(;b;z)=z^{-m}(2-b)_{m-1}\left ( \, _0 \tilde F_1(;b-m-1;z)\sum_{j=0}^{m-1}\left [ \frac{(m-j-1)!(-z)^j}{j!(m-2j-1)!(2-b)_j(b-m)_j} \right ]\\+(1+m-b)\,_0\tilde F_1(;b-m;z)\sum_{j=0}^{m-1}\left[\frac{(m-j)!(-z)^j}{j!(m-2j)!(2-b)_j(b-m-1)_j} \right ] \right ) \\ =z^{-(b-2)}(2-b)_{b-3}\left ( \, _0 \tilde F_1(;1;z)\sum_{j=0}^{b-3}\left [ \frac{(b-j-3)!(-z)^j}{j!(j+1)!(b-2j-3)!(2-b)_j} \right ]\\-\,_0\tilde F_1(;2;z)\sum_{j=0}^{b-3}\left[\frac{(b-2-j)!(-z)^j}{(j!)^2(b-2-2j)!(2-b)_j} \right ] \right ),\,b\geq 3$$