Not what you're really looking for, but here's a "word answer" in terms of a probabilistic interpretation.

Suppose Alice and Bob are playing a game where they are presented with $n$ indistinguishable bees and have to separate them into

• 1 queen bee,
• $n$ worker bees, and
• $n$ drones.

They both have to try to guess what groups the other has chosen. They play this game for increasing number of bees $n=1,2,3,\dots$

The probability that Alice is right is $$ \frac1{\binom{2n+1}{n}\binom{n}{1}} = \frac{(n!)^2}{(2n+1)!} = B(n+1,n+1) $$

Thus the number you're interested in is the expected number of times Alice and Bob both guess correctly.

For the record here is the usual derivation relating the integral to the infinite series in terms of the beta function $B$.

Let $a(x)=x(1-x)$ and $b(x,y)=a(x)a(y)$, and $$ c(x,y)=\frac1{1-b(x,y)} = \sum_{n=0}^\infty b(x,y)^n. $$ In terms of the beta function $B$ $$ \int_0^1 a(x)^n\,dx = \frac{(n!)^2}{(2n+1)!} = B(n+1,n+1) $$ so $$ \int_{[0,1]^2} c(x,y)\,dx\,dy = \sum_{n=0}^\infty B(n+1,n+1)^2 = \sum_{n=1}^\infty B(n,n)^2 = \sum_{n=1}^\infty \left(\frac{\Gamma(n)^2}{\Gamma(2n)}\right)^2 $$ $$ =\sum_{n=1}^\infty \frac{\Gamma(n)^4}{\Gamma(2n)^2} = {_3}F_2\left(1,1,1; \frac32,\frac32; \frac1{16}\right). $$ (Wolfram Alpha link for the last equation)

Not what you're really looking for, but here's a "word answer" in terms of a probabilistic interpretation.

Suppose Alice and Bob are playing a game where they are presented with $2n+1$ indistinguishable bees and have to separate them into

• 1 queen bee,
• $n$ worker bees, and
• $n$ drones.

They both have to try to guess what groups the other has chosen. They play this game for increasing $n=0,1,2,3,\dots$

The probability that Alice is right is $$ \frac1{\binom{2n+1}{n}\binom{n}{1}} = \frac{(n!)^2}{(2n+1)!} = B(n+1,n+1) $$

Thus the number you're interested in is the expected number of times Alice and Bob both guess correctly.

For the record here is the usual derivation relating the integral to the infinite series in terms of the beta function $B$.

Let $a(x)=x(1-x)$ and $b(x,y)=a(x)a(y)$, and $$ c(x,y)=\frac1{1-b(x,y)} = \sum_{n=0}^\infty b(x,y)^n. $$ In terms of the beta function $B$ $$ \int_0^1 a(x)^n\,dx = \frac{(n!)^2}{(2n+1)!} = B(n+1,n+1) $$ so $$ \int_{[0,1]^2} c(x,y)\,dx\,dy = \sum_{n=0}^\infty B(n+1,n+1)^2 = \sum_{n=1}^\infty B(n,n)^2 = \sum_{n=1}^\infty \left(\frac{\Gamma(n)^2}{\Gamma(2n)}\right)^2 $$ $$ =\sum_{n=1}^\infty \frac{\Gamma(n)^4}{\Gamma(2n)^2} = {_3}F_2\left(1,1,1; \frac32,\frac32; \frac1{16}\right). $$ (Wolfram Alpha link for the last equation)

This is a bit off topic, but I started thinking about a probabilistic interpretation of your constant. Here's the simplest I could come up with.

Suppose Alice and Bob are playing a game where they are presented with $n$ indistinguishable bees and have to separate them into

• 1 queen bee,
• $n$ worker bees, and
• $n$ drones.

They both have to try to guess what groups the other has chosen. They play this game for increasing number of bees $n=1,2,3,\dots$

The probability that Alice is right is $$ \frac1{\binom{2n+1}{n}\binom{n}{1}} = \frac{(n!)^2}{(2n+1)!} = B(n+1,n+1) $$

Thus the expected number of times Alice and Bob both guess correctly is the number you're interested in.

For the record here is the usual derivation relating the integral to the infinite series in terms of the beta function $B$.

Let $a(x)=x(1-x)$ and $b(x,y)=a(x)a(y)$, and $$ c(x,y)=\frac1{1-b(x,y)} = \sum_{n=0}^\infty b(x,y)^n. $$ In terms of the beta function $B$ $$ \int_0^1 a(x)^n\,dx = \frac{(n!)^2}{(2n+1)!} = B(n+1,n+1) $$ so $$ \int_{[0,1]^2} c(x,y)\,dx\,dy = \sum_{n=0}^\infty B(n+1,n+1)^2 = \sum_{n=1}^\infty B(n,n)^2 = \sum_{n=1}^\infty \left(\frac{\Gamma(n)^2}{\Gamma(2n)}\right)^2 $$ $$ =\sum_{n=1}^\infty \frac{\Gamma(n)^4}{\Gamma(2n)^2} = {_3}F_2\left(1,1,1; \frac32,\frac32; \frac1{16}\right). $$ (Wolfram Alpha link for the last equation)

Not what you're really looking for, but here's a "word answer" in terms of a probabilistic interpretation.

Suppose Alice and Bob are playing a game where they are presented with $n$ indistinguishable bees and have to separate them into

• 1 queen bee,
• $n$ worker bees, and
• $n$ drones.

They both have to try to guess what groups the other has chosen. They play this game for increasing number of bees $n=1,2,3,\dots$

The probability that Alice is right is $$ \frac1{\binom{2n+1}{n}\binom{n}{1}} = \frac{(n!)^2}{(2n+1)!} = B(n+1,n+1) $$

Thus the number you're interested in is the expected number of times Alice and Bob both guess correctly.

For the record here is the usual derivation relating the integral to the infinite series in terms of the beta function $B$.

Let $a(x)=x(1-x)$ and $b(x,y)=a(x)a(y)$, and $$ c(x,y)=\frac1{1-b(x,y)} = \sum_{n=0}^\infty b(x,y)^n. $$ In terms of the beta function $B$ $$ \int_0^1 a(x)^n\,dx = \frac{(n!)^2}{(2n+1)!} = B(n+1,n+1) $$ so $$ \int_{[0,1]^2} c(x,y)\,dx\,dy = \sum_{n=0}^\infty B(n+1,n+1)^2 = \sum_{n=1}^\infty B(n,n)^2 = \sum_{n=1}^\infty \left(\frac{\Gamma(n)^2}{\Gamma(2n)}\right)^2 $$ $$ =\sum_{n=1}^\infty \frac{\Gamma(n)^4}{\Gamma(2n)^2} = {_3}F_2\left(1,1,1; \frac32,\frac32; \frac1{16}\right). $$ (Wolfram Alpha link for the last equation)