Interpreting a result from the "inverse symbolic calculator" - hypergeometric function I punched the following number, $1.7820738631661201045831848697933955378\ldots$ into the classic look-up of the inverse symbolic calculator. Admittedly the result returned from the calculator are not clear to me at all.  
First it reads:

F(a,b;z) with a,b in F12 and z in F60: F(a,b;z) hypergeometric
  function 1782073863166120=F(1,1;26/45)

and it also reads

F(a,b;z) with a,b in F12 and z in F60: F(a,b;z) hypergeometric
  function 1782073863166120=F(1/3,1/3;26/45)

How should I be interpreting the results from the ISC?
I did read Hypergeometric Functions to try and start to begin having an idea what hypergeometric functions are. 
 A: 
I was thinking possibly: here

You are right. It is the Confluent Hypergeometric function of the first kind. 
$$_1F_1(a,b;z)=\sum_{k=0}^{\infty}\frac{(a)_k}{(b)_k}\cdot \frac{z^k}{k!}=1+\frac{a}b+\frac{a(a+1)}{b(b+1)}\cdot \frac{z^2}{2!}+\frac{a(a+1)(a+2)}{b(b+1)(b+2)}\cdot \frac{z^3}{3!}+\ldots$$
For $a=b$ the fraction $\frac{(a)_k}{(b)_k}$ is equal to $1$. Therefore
$_1F_1\left(1,1;z\right)=_1F_1\left(\frac13,\frac13;z\right)=\sum_{k=0}^{\infty} \frac{z^k}{k!}$
This is the series expansion of $e^z$. Consequently for $z=\frac{26}{45}$
$_1F_1\left(1,1;\frac{26}{45}\right)=_1F_1\left(\frac13,\frac13;\frac{26}{45}\right)=e^{\frac{26}{45}}=1.7820738631661201045831848697933955378...$
A: If I understand it correctly, the ISC index shows what types of expressions were used to build the ISC database. And F(a,b;z) with a,b in F12 and z in F60: F(a,b;z) hypergeometric function is one of the entries there.
As explained in the index, F12 is the set of Farey fractions with denominator 12 or less. And similarly for F60, I presume.
So your search must have generated a couple of hits from the table made by evaluating $F(a,b;z)$ for all $a$ and $b$ in the set F12 and all $z$ from the set F60.
Here, F(a,b;z) seems to be the hypergeometric function of type ${}_2F_0(a,b;z)$, althought it's not completely clear to me. It might also be ${}_1F_1(a;b;z)$, but the comma (rather than a semicolon) between a and b makes the first option seem more likely.
