Is the “almost-identity” $\sum_{k=0}^\infty \left[\pi^{\frac k2}\big/\Gamma{\left(\frac k2+1\right)}\right]\approx46$ significant or a coincidence?

Question

Recently, I read an article about “almost-identities”. It said, that for every “almost-identity” we have to decide whether it is a coincidence or not. By myself, I discovered that $$ \sum_{k=0}^\infty \frac{\pi^{\frac k2}}{\Gamma{\left(\frac k2+1\right)}}=e^{\pi}\left(1+\text{erf}\left(\sqrt\pi\right)\right)\approx45.9993260894... $$ which is surprisingly close to $46$. So my question is: is this a mere coincidence or can it be “proven” in some sense?

Answer 1

It is interesting to note that

$$\sum _{k=0}^{\infty }{\frac {{\pi }^{1/2\,k}}{\Gamma \left( 1/2\,k+a \right) }}={\frac {{{\rm e}^{\pi }} \left( \Gamma \left( a \right) - \Gamma \left( a-1,\pi \right) a+\Gamma \left( a-1,\pi \right) \right) }{\Gamma \left( a \right) {\pi }^{a-1}}}+{\frac {\sqrt {\pi } \left( 2\,{{\rm e}^{\pi }}\Gamma \left( 1/2+a \right) - \left( 2\,a-1 \right) {{\rm e}^{\pi }}\Gamma \left( -1/2+a,\pi \right) \right) }{2\Gamma \left( 1/2+a \right) {\pi }^{-1/2+a}}} $$

Then it is possible to solve the equation

$${\frac {{{\rm e}^{\pi }} \left( \Gamma \left( a \right) - \Gamma \left( a-1,\pi \right) a+\Gamma \left( a-1,\pi \right) \right) }{\Gamma \left( a \right) {\pi }^{a-1}}}+{\frac {\sqrt {\pi } \left( 2\,{{\rm e}^{\pi }}\Gamma \left( 1/2+a \right) - \left( 2\,a-1 \right) {{\rm e}^{\pi }}\Gamma \left( -1/2+a,\pi \right) \right) }{2\Gamma \left( 1/2+a \right) {\pi }^{-1/2+a}}} = 46$$

and the solution is

$$a=0.999987485786729129363813640894$$

which is very near to $1$. According with this the result of @user109899 it is not a coincidence.

Answer 2

If you consider the level of simplicity of the expression, you will find that there are at least 10,000 equally simple expressions that you might equally well have checked. For example, instead of $e^\pi$ you might have taken $2^\pi$ or $3^e$ or $\sqrt{2}^{\sqrt{3}}$ or .... If we allow he all combinations using 2 symbols among the numbers from 1 to 10, the constants $e$, $\pi$, and $\gamma$ possibly multiplying those numbers, and square roots (after all, you felt comfortable with a square root in the argument of erf), that one first term has about 50 possibilities.

Similarly, the $\text{erf}(\sqrt{\pi})$ expression could have used any of at least 20 functions about as familiar as erf, and maybe 20 arguments of comparable complexity to the one here.

In the end, the fact that this matches an integer to 3 decimal places is on that basis explainable as a coincidence.

There are other near-equalities that are, in fact, not coincidences. Some are encountered/explained when you study class numbers. But most of those equalities have an astonishing degree of closeness, not just 3 digits.