Given the definition for the $H$ function in terms of the of a hypergeometric function and knowing that $n$ and $k$ are always positive natural numbers:
$$ H(n+k)={_3F_2}\left({1,n+(1+2k)/2,n+(1+2k)/2 \atop n+k,n+k};1/2\right) $$
My problem has to do with the following expression $Z$:
$$ Z=(a\,H(n+2)+b\, H(n+1)+c\, H(n+1)\,H(n+2))\, T(n)+(x\,H(n+1)+y\, H(n)+z\, H(n-1))\,T(n-1) $$
$a,b,c,x,y,z$ are all closed form expressions in terms of $n$ that are very big and are not needed to solve my problem.
$T$ is an ascending function that is positive for positive input. I'm at a point where the function $T$ can be any function and $Z$ doesn't change in value. My goal is to define $T$ in such a way that $Z$ can be simplified to be a closed form expression (no hypergeometric functions) $$T:\Bbb R\longrightarrow\Bbb R$$
To do that, I would need to express $H(n+1), H(n+2)$ and $H(n-1)$ in terms of $H(n)$.
$$ H(n+2)=p\, H(n) $$
$$ H(n+1)=r\, H(n) $$
$$ H(n-1)=u\, H(n) $$
Where $p,r,u$ are closed form expressions. That way I can factor out $H(n)$ and have the following formula for $Z$:
$$ Z=(ap+br+cpr\, H(n))\,H(n)\,T(n)+(xr+y+zu)\, H(n)\,T(n-1) $$
$$ Z=H(n)\,((ap+br+cpr\, H(n))\,T(n)+(xr+y+zu)\,T(n-1)) $$
And then I could find which function $T$ needs to be in order to simply the $H(n)$ in front and be left with a closed form expression.
Since there is no closed form for $H(n+k)$, my goal is to find a formula that allows me to transform any H(n+k) into a H(n) multiplied by some closed form expression. Note that this is just one idea. I would also accept a formula for H(n) that depends on 2 terms, $H(n-k)$ and $H(n-j)$, or 3 terms or 4.
My guess is that I need to obtain it with a combination of the contiguous formulas found on the link below, but since there are so many combinations, I'm wondering if someone already found the formula.
https://functions.wolfram.com/HypergeometricFunctions/Hypergeometric3F2/17/ShowAll.html
